Introduction to Projectile Motion - Formulas and Equations

Summary

This video covers essential kinematic equations used to solve projectile motion problems. It explains the four key equations for motion with constant acceleration and delves into different projectile trajectories: horizontal launches, angled launches, and launches from elevated positions. The video demonstrates how to calculate displacement, velocity, time, range, and angle for each scenario, utilizing equations such as d = v_0t + (1/2)at^2 and the range formula R = (v^2 * sin(2θ)) / g. Viewers also learn how to avoid common mistakes in calculations and how to apply these equations effectively.

Takeaways

• 📐 The displacement of an object moving at constant speed is equal to velocity multiplied by time.
• 🚀 For constant acceleration, the final speed is equal to the initial speed plus acceleration times time.
• 📊 The square of the final speed is equal to the square of the initial speed plus two times acceleration multiplied by displacement.
• 🕰️ Displacement can be found using the average speed (initial speed plus final speed divided by two) multiplied by time.
• 📏 Displacement (D) is equal to initial velocity times time plus half the acceleration times the square of time.
• 📉 In projectile motion, separate calculations for the horizontal (X) and vertical (Y) directions are needed.
• 🟰 In horizontal projectile motion, the height (H) can be calculated using H = 1/2 * acceleration * time^2.
• 🧮 To find the range of a projectile, use the equation Range = velocity * time in the X direction.
• 🎯 The total time of flight for a projectile launched at an angle is 2 * V * sin(Theta) / g.
• 📐 The maximum height of a projectile is given by the formula Height = (V^2 * sin^2(Theta)) / (2 * g).

Q & A

What are the basic kinematic equations for constant speed motion?

For constant speed motion, the basic kinematic equation is displacement equals velocity multiplied by time.

List the four key equations for motion with constant acceleration.

The four key equations for motion with constant acceleration are: 1) final speed equals initial speed plus acceleration times time, 2) the square of the final speed equals the square of the initial speed plus 2 times acceleration times displacement, 3) displacement equals average speed times time (where average speed is the sum of initial and final speeds divided by two), and 4) displacement equals initial velocity times time plus half of acceleration times the square of time.

What is the difference between displacement and distance?

Displacement and distance are the same if an object moves in one direction without changing direction. If the object changes direction, displacement and distance are not the same. Displacement is the straight-line distance from the initial to the final position, while distance is the total path length traveled.

What are the three types of trajectories in projectile motion?

The three types of trajectories in projectile motion are: 1) horizontal launch from a height (like off a cliff), 2) launched at an angle from ground level, and 3) launched at an angle from an elevated position.

How can you calculate the height of a cliff if an object is projected horizontally and falls off?

To calculate the height of a cliff, use the equation H equals 1/2 times acceleration due to gravity times time squared (H = 1/2 * g * t^2), where time is the time it takes for the object to hit the ground.

What is the formula to calculate the range of a projectile launched horizontally from a cliff?

The formula to calculate the range (horizontal distance traveled) of a projectile launched horizontally from a cliff is range equals initial horizontal velocity times time (R = Vx * t).

How do you find the speed of a ball just before it hits the ground in projectile motion?

To find the speed of a ball just before it hits the ground, you need to calculate both the horizontal velocity (which remains constant) and the vertical velocity using the equation VY final equals initial vertical velocity plus acceleration due to gravity times time.

What is the relationship between the angle of launch, initial velocity, and maximum height for a projectile launched at an angle from the ground?

The maximum height of a projectile launched at an angle from the ground is given by the equation H equals initial velocity squared times sin^2(angle) divided by 2 times gravity (H = V^2 * sin^2(Theta) / 2G).

How can you determine the time it takes for a projectile to hit the ground when launched at an angle from an elevated position?

The time it takes for a projectile to hit the ground when launched at an angle from an elevated position can be found using the quadratic formula to solve the equation Y final equals initial height plus initial vertical velocity times time plus half of gravity times time squared, set to zero since Y final is ground level.

What is the equation to calculate the range of a projectile launched at an angle from an elevated position?

The range of a projectile launched at an angle from an elevated position is calculated using the equation range equals initial horizontal velocity times time (R = Vx * t), where time is the time it takes to hit the ground and initial horizontal velocity is the horizontal component of the initial velocity (Vx = V * cos(Theta)).

Outlines

00:00

📚 Basic Kinematic Equations for Constant Speed and Acceleration

This paragraph introduces the fundamental kinematic equations necessary for solving projectile motion problems. It starts with the basic equation for constant speed, where displacement equals velocity times time. Then, it moves on to four key equations for motion with constant acceleration: the final velocity formula, the equation relating the square of the final velocity to the initial velocity and acceleration, the average velocity formula, and the displacement formula involving initial velocity, acceleration, and time. The paragraph also clarifies the difference between displacement and distance, emphasizing that displacement is the straight-line distance from the initial to the final position, while distance is the total path length traveled. The distinction is crucial when the direction of motion changes.

05:00

🌟 Projectile Motion Trajectories and Equations

The second paragraph delves into the specifics of projectile motion, starting with the scenario where an object is projected horizontally from a cliff. It introduces the equation H = 1/2 * a * t^2 to calculate the height of the cliff based on time. The discussion then separates the motion into horizontal (x-direction) and vertical (y-direction) components, using the equations D = Vx*t and H = 1/2 * a*t^2, where D is the range, and H is the height. The paragraph also covers how to find the speed of the ball just before it hits the ground, using the horizontal and vertical velocity components and the inverse tangent function to determine the angle of impact.

10:01

🔍 Launching at an Angle: Time and Maximum Height Equations

The third paragraph explores the trajectory of a ball launched at an angle from the ground, discussing the time it takes to reach the maximum height (point B) and the total time to return to the ground (point C). It introduces the equation t = V*sin(Theta)/G to calculate the time to reach the maximum height and explains that the total time for the ball to hit the ground is twice this value due to the symmetry of the trajectory. The paragraph also provides the formula for calculating the maximum height, H = V^2 * sin(2*Theta) / 2G, and discusses the range equation R = V^2 * sin(2*Theta) / G, which applies to symmetrical trajectories.

15:01

📉 Projectile from a Height: Time and Range Calculations

This paragraph addresses the scenario where a ball is launched at an angle from a height, such as a cliff. It explains how to calculate the time it takes for the ball to hit the ground, using the quadratic formula to solve for time when the final y-position is zero. The paragraph also discusses an alternative method to find the time without using the quadratic formula, by first calculating the time to reach the maximum height and then using it to find the total time to hit the ground. Additionally, it covers the range calculation for this type of trajectory, emphasizing the correct use of the range equation based on the trajectory's symmetry.

20:02

🎯 Final Velocity and Angle Before Impact

The fifth paragraph focuses on calculating the final velocity and angle of a ball just before it hits the ground after being launched from a height. It explains that the horizontal velocity (VX) remains constant, and the vertical velocity (VY) can be found using the initial vertical velocity and acceleration due to gravity. The final velocity is then calculated using the horizontal and vertical components, and the angle is determined using the inverse tangent function. The paragraph also discusses how to describe the angle relative to the horizontal or the positive x-axis, highlighting the importance of understanding the problem's requirements.

25:04

🔚 Summary of Projectile Motion Equations

The final paragraph summarizes the key equations for projectile motion, categorizing them based on the type of trajectory discussed in the previous paragraphs. It reviews the equations for calculating height and range when an object falls from a height, the time and range equations for a ball launched at an angle from the ground, and the maximum height and range equations for a ball launched at an angle from a height. The paragraph also reiterates the importance of using the correct equations based on the specific details of the projectile motion problem.

Keywords

💡Projectile Motion

Projectile motion refers to the motion of an object thrown or projected into the air, where it moves along a curved path under the action of gravity alone. In the video, this concept is central as it discusses equations and principles necessary to solve problems involving objects in projectile motion. For example, the video covers different scenarios such as an object thrown horizontally from a cliff and how to calculate its trajectory and landing point.

💡Kinematic Equations

Kinematic equations are fundamental principles in physics that describe the motion of an object. The video reviews these equations, which are essential for understanding how to calculate displacement, velocity, and acceleration in various scenarios, including projectile motion. The equations discussed include those for constant speed and constant acceleration, which are used to derive more specific equations for projectile motion.

💡Displacement

Displacement is the change in position of an object. In the video, displacement is highlighted as a key concept, particularly in distinguishing between horizontal (x-direction) and vertical (y-direction) movements in projectile motion. The script explains how displacement can be calculated using different kinematic equations depending on whether the motion involves constant speed or constant acceleration.

💡Constant Acceleration

Constant acceleration implies that the rate of change of velocity is constant over time. The video emphasizes the importance of understanding constant acceleration, especially in the context of gravitational acceleration (g=9.8 m/s²), which affects the vertical component of projectile motion. Equations involving constant acceleration are used to calculate the final velocity and displacement in projectile problems.

💡Gravitational Acceleration

Gravitational acceleration, often denoted as 'g', is the acceleration experienced by an object in free fall due to gravity. In the video, gravitational acceleration is a critical factor in calculating the vertical component of projectile motion. It is used in equations to determine how the vertical velocity and displacement change over time.

💡Trajectory

Trajectory describes the path that a projectile follows through the air. The video discusses different types of trajectories, such as those that start from a cliff or from the ground at an angle, and how to calculate various aspects of these trajectories using kinematic equations. Understanding trajectory is crucial for predicting the motion and impact points of projectiles.

💡Horizontal and Vertical Components

In projectile motion, the motion is typically broken down into horizontal (x-direction) and vertical (y-direction) components. The video explains how to separate these components to apply kinematic equations correctly. For instance, horizontal motion at a constant speed is independent of the vertical motion, which is affected by gravity.

💡Range

Range, in the context of projectile motion, refers to the horizontal distance traveled by an object before it lands. The video provides equations to calculate the range based on the initial velocity and the angle of projection. Understanding how to calculate range is essential for solving problems involving the distance an object travels horizontally.

💡Time of Flight

Time of flight is the total time an object spends in motion from the moment it is projected until it hits the ground. The video script discusses how to calculate the time of flight for different types of projectile motion scenarios, which is crucial for determining when and where the object will land.

💡Quadratic Equation

A quadratic equation is a second-degree polynomial equation. In the video, the quadratic equation is mentioned in the context of finding the time it takes for an object to hit the ground when launched from a height, which involves solving for time 't' in an equation derived from kinematic principles.

💡Angle of Projection

The angle of projection is the angle at which a projectile is launched relative to the horizontal. The video explains how the angle of projection affects the trajectory and range of the projectile. It shows how to calculate various aspects of the motion, such as time to reach maximum height and range, using the angle of projection.

Highlights

Basic kinematic equations for constant speed motion are reviewed.

Four key equations for motion with constant acceleration are introduced.

The distinction between displacement and distance is explained.

Three types of projectile motion trajectories are outlined.

Equation for calculating the height of a cliff in projectile motion is presented.

Separation of motion into x and y directions for projectile motion is discussed.

Calculation of the range in projectile motion is detailed.

How to find the speed of a ball just before it hits the ground in projectile motion is explained.

The relationship between horizontal and vertical velocity components is described.

Derivation of the time to reach maximum height in an angled trajectory is shown.

Total time for a ball to travel from launch to landing in an angled trajectory is calculated.

Derivation of the equation for maximum height in an angled trajectory is provided.

Calculation of the range in an angled trajectory is explained.

How to find the speed and angle just before a ball hits the ground is detailed.

Different ways to describe the angle of velocity relative to the horizontal are discussed.

Equations for projectile motion from an elevated position are summarized.

The importance of using the correct equations for different projectile motion scenarios is emphasized.

Practical applications of projectile motion equations are highlighted.

Transcripts

in this video we're going to go over

some equations that you need to know to

solve projectile motion

problems so let's review some basic

kinematic equations whenever an object

is moving with constant speed

displacement is equal to Velocity

multiplied by time now when an object is

moving with constant acceleration there

are four equations you need to know the

final speed is equal to the initial

speed plus the product of the

acceleration and the

time there's also this equation the

square of the final speed is equal to

the square of the initial speed plus 2 *

the product of the acceleration and the

displacement

displacement is equal to the average

speed

which the average speed is basically the

sum of the initial speed and the final

speed divided by two multiplied by the

time

as you can see the third equation that I

listed here looks like the first

equation for a constant

speed okay I did not want to do that so

I can rewrite this equation like

this V average times T where V average

is initial plus final ID two now there's

another equation that you need to know D

is equal to V initial

t plus 12 a t^2 so make sure you know

that equation too so what exactly is D

you can use d as displacement or

distance distance and displacement are

the same if an object moves in One

Direction and doesn't change direction

if it changes Direction then

displacement and distance is not the

same but technically D is

displacement

displacement is basically the difference

between the final position minus the

initial position

now you can use displacement in the X

direction or you can use it in the y

direction so just keep that in

mind now there's three types of shapes

for projectile motions that you need to

be familiar

with let's say

if we have an

object and it comes off a cliff

horizontally and then eventually hits

the ground that's the first type of

trajectory you're going to be dealing

with so what equations do we need for

this

situation the first equation that's

going to help you is this H is equal to

12 a

t^2 where H represents the height of the

cliff and T is the time it takes to hit

the

ground this equation comes from this

equation D is equal to V initial

t plus

12

a^2 now when you're using these

equations you need to ask yourself am I

dealing with the X direction or the y

direction you got to separate X and Y

for projectile motion

problems we're going to apply this

equation along the Y Direction so this

is really

Dy is equal to VY

initial time

t

plus 12 a y * t

squ Dy the displacement in the y

direction is basically the

height so that's

H VY at the top is always zero because

the object is moving horizontally the

initial speed of the ball is VX not

VY and so we get this equation so H is

equal to 12 at

squ so let me just redraw this

picture so make sure you know the first

equation that we just

mentioned now in addition to knowing

this equation which if you know the time

you could find the height of the cliff

there's also another one we know that D

is equal to

VT this is the range of the graph the

distance between where the ball lands

and the base of the

cliff now if we apply this equation in

the X Direction the displacement in the

X direction is the range so we can say

that the range is equal to

vxt and for this type of trajectory VX

is the initial speed of the ball so

that's how you can calculate the range

if you have the range you can find the

time and then using the time you could

find the

height so these are the two main

equations that you'll need

for this particular trajectory now

there's some other equations sometimes

you got to find the speed of the ball

just before it hits the

ground whatever VX was at the beginning

VX would be the same for projectile

motion problems VX does not change the

acceleration in the X direction is zero

so VX is constant the acceleration in

the Y Direction that's the gravitational

acceleration it's

9.8 and so v y Chang

es to calculate VY we can use this

equation V final equals V

initial plus a t but we're going to use

it in the y

direction so we know that VY initial is

zero at the top VY is always zero so you

could find VY final using this equation

now to find the speed of the ball just

before hits the ground

you need to use

the horizontal velocity and the vertical

velocity and if you need to find the

angle you can use this equation it's

inverse tangent VY /

VX now the second type of

trajectory involves this type of

fixtion so let's say if we have a ball

and we kick the ball from the ground it

goes up and then it goes

down so that ball is going to have an

angle Theta relative to the horizontal

and it's going to be launched at a speed

V this is not VX or v y this is

V now let's draw the vector v so here's

V and here's the angle

Theta V has an X component and it has a

y component the X component is VX the Y

component is

VY so VX is equal to V cosine

Theta and VY is equal to V sin

Theta and v as you mentioned before is

the square root of vx2 plus v y^2 and if

you want to find Theta it's inverse

tangent VY over

VX now let's define this as point a

point B and point

C what is the equation that we need to

calculate the time it takes to go from A

to B or to reach the maximum height

which is at position

B how long does it take to go from A to

B now to derive that

equation we need to start with this

equation V final equals V initial plus a

but we're going to use it in the y

direction so this is going to be VY

final equals VY initial plus

a at the top that is that position B VY

is always zero because it's not going up

anymore so B is the final position a is

initial position so VY final is

zero v y initial is basically V sin

thet as we uh wrote it in this equation

acceleration and the y direction is

basically

G so solving for T we need to move this

term to the other side so negative V sin

Theta is equal to GT and dividing both

sides by G we can see that t is equal to

V sin

Theta over

G so that's the time it takes to go from

A to B the time it takes to go from a to

c is twice the value notice that the

graph is symmetrical so if it takes 5

Seconds to go from A to B it takes 5

Seconds to go from B to C and so to go

from a to c it's 10 seconds so the total

time is just 2 V sin Theta / G if you're

wondering what happen to the negative

sign know that g is 9.8 so it cancels

with the negative sign but if you plug

in positive 9.8 you don't need the

negative sign anymore either case time

is always positive so so just make sure

you report a positive value for

time

now let's talk about some other

equations that relate to this

shape how can we derive an expression to

calculate the maximite between position

a and position

B so going from A to

B let's use this particular kinematic

equation V final or VY final squ is

equal to VY initial

squ plus 2 * a y *

Dy so you've seen it as V final s equals

V initial

S Plus 2 a d I simply added the

subscript y to it because the height is

in the y

direction now at the top VY is equal to

zero so position B is the final position

so VY final is zero VY initial we know

that VY is V sin Theta so vy^ 2 is V sin

Theta 2 and of course we have plus two

times the acceleration in the y

direction is G the displacement in the y

direction is the same as the height so

we can put H at this

point so let's move this term to the

left

so V ^2 sin 2 thet if you distribute the

two is equal to 2 GH to now let's divide

by

2G so the maximum height if we ignore

the negative sign is V

^2 sin s thet over

2G so that's how you can find the height

if you have the angle

Theta and the velocity

not VX or v y but the velocity

V now what equation can we come up with

in order to

calculate the

range how can we derive an equation to

find a range of the

graph what equation would you use the

range is the horizontal distance

or the horizontal displacement of the

ball now we said that the range is equal

to

vxt and you could find VX by using the

fact that VX is V cosine

Theta now the

range is the horizontal displacement

from points a to

c so what is the time from point a to c

as you mentioned earlier in is video the

time it takes for the ball to go from a

to c is equal to 2 V sin Theta /

G so we have V cosine Theta and we're

going to replace t with 2 V sin Theta

over

G now V * V is basically

v^2 so we have v^2 time 2 sin

Theta cine

Theta /

G now there's something called a double

angle formula in

trigonometry and here it

is sin 2

thet is equal to 2 sin Theta cosine

Theta so therefore we can replace the

expression 2 sin Theta cosine Theta with

sin 2

thet and so therefore the

range is v^

2 sin 2 Theta /

G so that's how you can derive an

equation for the range if you have this

type of

trajectory now the third type of

trajectory that you're going to see

involves

a ball being launched at an angle from a

cliff or from some elevated position

above ground level so it's going to go

up and then it's going to go

down so H is the height of the cliff and

R once again is the range of the

ball let's call this

uh position a position B and position

C so one of the first type of questions

that you might find with this type of

problem is calculating the time it takes

to hit the ground going from a to c now

this graph is not the same as this

trajectory where the time it takes to go

from a to c is 2 V sin Theta over G if

you use this equation that'll give you

the time it takes to get to this

position which is symmetrical to a so

don't do it it's not going to

work so therefore we need to do

something

else

it turns out that there's another way to

calculate the time I'm going to show you

two ways so please be

patient let's start with this equation

displacement is equal to V initial t

plus 12 a t^2 but we're going to apply

it in the y

direction displacement in the y

direction is the difference between the

final position minus initial

position and then V initial but in the y

direction is VY

initial acceleration in the y direction

is basically 9.8 or

G so moving this term to the other side

perhaps you've seen this equation Y

final equals y initial y initial is

basically the height of the

cliff plus VY initial T remember VY

initial is V sin Theta Theta is the

angle above the horizontal plus 12

GT2 now to find the time it takes to hit

the ground you need to realize that the

position the Y value at ground level is

zero so if you replace it with

zero replace y initia with h replace v y

initia with v sin

Theta you now have everything you need

to solve for T you know what G is

however you need to use a quadratic

equation you want to make sure

everything's on one side and on the

other side you have a zero so using the

quadratic formula T is equal to B plus

or minus < TK B ^2 - 4 a c ID 2

a so let's say if you get an equation

that looks like this and you put it in

standard form 4.9

t^2 Plus 8 T

plus 100 or something like

that in this case well actually this is

going to be 4.9 t^2 you got to make sure

you plug Inga 9.8 for g a is 4.9 B is 8

C is 100 and then just plug it into this

formula and you should get the

answer now let's go back to the same

trajectory

now what's another way in which we can

calculate the

time that is the time it takes to hit

the ground to go from position a to

position

C Is there a way in which we can get the

same answer without using the quadratic

equation it turns out that there is to

go from A to B notice that it's the same

as this trajectory

so we can use the equation that we used

in that

trajectory the time it takes to go from

A to

B is

simply V sin Theta / G and typically in

this problem you'll be given

V that is just the speed of the

ball and you're going to be given uh

Theta the angle relative to the

horizontal now how can we find the time

it takes to go from B to

C well in order to do that you need to

find the height between positions a and

position

B to find that height it's going to be

equal

to v^ 2 sin 2 and I believe it was over

2G if I'm not

mistaken I believe that's the equation

once you get the height then the total

height which let's call the total

height y

Max the total height from B to C we can

use that to find a time now you've seen

this equation H is equal to 12 a^2 but

in this particular situation between B

and C

H it's really the sum of a h plus

Capital H so that's the total height

from B to

C and that's equal to 12 a which is the

same as G time t^2 so basically if you

rearrange the

equation T going from B to

C is going to

be let's call this y Max so

2 * y

maxide div G sare root that's going to

give you the time it takes to go from B

to C so the total time is simply the sum

of these two

values and that's how you can avoid

using the quadratic formula if you don't

want

to now how can we find the

range what equation would you use to

calculate the range of the

ball

to find a range use this equation

vxt for this trajectory do not use the

equation v^2 sin 2 Theta over G only use

this equation if you have a symmetrical

trajectory this is the only time you

should use this equation otherwise if

you use it for this problem you're going

to get the range

between these two points

let's let's call this point a and point

D you're going to get the range between

those two points and you don't want

that so to find a range for this type of

trajectory you use the equation VX *

T now keep in mind

VX is equal to V cosine

Theta so the range is simply V * cosine

thet * T you can use this form of the

equation if you want to but you need to

find the time it takes to go from a to c

before you can use it which using the

last two equations you can get that

answer

now and that's it that's all you need to

do to find the range now sometimes you

might be asked to find the speed of the

ball just before it hits the

ground so first you need to realize that

VX is constant so whatever VX you have

here it's going to be the same at Point

C at point a point B and point C VX is

the

same so let me make sure I write that VX

is constant it does not

change so whatever VX you have here just

we're going to use that to find the

final speed of the ball just before it's

the

ground now just like last time we

mentioned earlier in this video we got

to find the vertical velocity as well

using this

equation so you need to find VY

final

and you know VY initial is basically V

sin Theta where V is the initial

velocity at point

A and G make sure you plug in negative

9.8 T this is the time it takes to go

from a to position C so use the total

time so this will give you the final

Vertical Velocity so once you have VX

and v y at Point C to find the final

speed just before hits the ground you

can now use this

equation and if you need to find the

angle as mentioned before it's going to

be inverse tangent VY / VX now let's

talk about the

angle so let's say this is the ball and

it's moving in this direction just

before it hits the

ground and let's say this is the x

axis and this is the Y

AIS making the ball the center

so the ball has an X component VX and it

has a y component VY you're looking for

V which is the velocity and the

magnitude of velocity is the

speed so once you use inverse tangent

Theta make sure you plug in positive

values for v y and VX even though v y is

going to be negative don't plug in a

negative value plug in a positive value

that will give you the reference angle

which is an angle between 0 and

90 now sometimes you might describe your

answer as being below the horizontal or

relative to the positive

xaxis so let's say

if the

angle let's say it's

60° so that's going to be the reference

angle so you can describe it as being

60°

below the horizontal which is the xaxis

or you could say it's uh positive

300 relative to the positive

x-axis so you have two ways of

describing the

angle just be careful um to get it the

right way in terms of the way the

problem wants you to describe it so it

could be 60 it could be 300 just think

about what they're asking for if it's

below the horizontal 60 is it if it's

measured from the positive xaxis then

it's 300 it's 360 minus 60 which is

300 so make sure you understand all of

the equations and when to use them so

just to

review for this type of trajectory

where the ball simply falls down it

travels horizontally from a cliff and

just falls off the height is equal to 12

a^2 and range is equal to vxt

and for all trajectories you can use

this

equation if you need to find the angle

use

this just plug in positive values for v

y and VX and also if you need to find VY

final to use it

here use this

equation now for the second

Dory these are the main equations that

you need just to review what we went

over so the time it takes going from

let's say A to B remember it's just V

sin

Theta / G and the time it takes to go

from a to c is twice that value it's 2 V

sin Theta over

G and the

range is V ^ 2 sin squ

divid

by let's see if I remember this it was

uh no no that's not the range that's the

height it was V sin 2 thet / G that's

what it

was and now for the other

one to find the maximum height it's V

^2 sin 2 /

2G so those are the four main equations

you need for this trajectory and also if

you need to find the angle just before

it hits the ground at position C is the

same as the angle when it uh left the

ground so and the speed at which it left

the ground is the same as the speed at

which it hits the

ground so let's say if it left the

ground at 20

m/s the speed before it hits the ground

is 20 since this trajectory is

symmetrical

A and C are the same the Ang going to be

the same and the speed will be the

same and then finally for the last

trajectory all of the equations that

you've seen for the first two you can

apply it to this

one the only thing that's different is

this equation Y final equals y initial

plus v y initial

t plus 12 a t^2

and so we're going to stop here so those

are all of the equations that you'll

need for these type of projectile motion

problems so that's it for this video and

thanks for watching