How To Solve Projectile Motion Problems In Physics

Summary

This video provides an in-depth explanation of projectile motion, focusing on key concepts like kinematic equations, constant speed, and acceleration. It covers the fundamental equations for displacement, velocity, and time, emphasizing the relationship between horizontal and vertical velocities in projectile motion. The instructor explains different types of trajectories, such as horizontal motion off a cliff and motion at an angle, highlighting important equations for calculating time, height, and range. The video also demonstrates solving example problems to apply these concepts, making it a comprehensive resource for understanding projectile motion in physics.

Takeaways

• 🚀 The kinematic equations at constant speed include displacement (d) = velocity (v) multiplied by time (t), and at constant acceleration, displacement equals average velocity times time.
• ⚡ Average velocity under constant acceleration is the sum of initial and final velocity, divided by two.
• 🎯 Key equations: final velocity (v final) = initial velocity (v initial) + acceleration (a) × time (t), and final velocity squared = initial velocity squared + 2 × acceleration × displacement.
• 🌍 Gravitational acceleration is approximated as -9.8 m/s², and in the example, it's rounded to -10 m/s² for simplicity.
• ⚖ Horizontal velocity (vx) remains constant in projectile motion, while vertical velocity (vy) changes due to gravitational acceleration.
• 💡 In a projectile, only gravity acts on the object after release; horizontal acceleration is zero, and vertical acceleration is constant.
• 🔄 The speed at the same height on both sides of the projectile’s arc is the same, but vertical velocity changes in direction (positive on the way up, negative on the way down).
• 📏 Three common projectile trajectories: (1) horizontal motion off a cliff, (2) motion starting from the ground going up and coming down, and (3) motion from a height kicked at an angle.
• 📐 The range (horizontal distance) in projectile motion can be calculated using v_x × time, while the height or vertical displacement follows the equation y_final = y_initial + v_initial × time + 1/2 × acceleration × time².
• 🧮 To find the final speed before an object hits the ground, use the Pythagorean theorem: v = sqrt(vx² + vy²).

Q & A

What are the basic kinematic equations under constant speed?

Under constant speed, displacement (d) is equal to velocity (v) multiplied by time (t): d = vt.

How do you calculate displacement under constant acceleration?

Displacement is equal to the average velocity multiplied by time. Average velocity is the sum of the initial and final velocity, divided by two.

What is the formula for final velocity under constant acceleration?

The final velocity (v_final) is equal to the initial velocity (v_initial) plus the acceleration (a) multiplied by time (t): v_final = v_initial + at.

How do you calculate the displacement if the acceleration is constant?

The displacement (d) can be calculated using the formula: d = v_initial * t + 0.5 * a * t^2.

What happens to the horizontal and vertical velocities of a projectile one second after launch?

The horizontal velocity remains constant, while the vertical velocity decreases by 10 m/s due to gravitational acceleration (assuming g = 10 m/s² for simplicity).

Why does the vertical velocity change but the horizontal velocity stays the same in projectile motion?

The vertical velocity changes due to gravitational acceleration acting in the vertical direction, while the horizontal velocity remains constant because there is no acceleration in the horizontal direction (neglecting air resistance).

How does speed differ from velocity?

Speed is the magnitude of velocity and is always positive. Velocity is a vector, meaning it has both magnitude and direction, and can be positive or negative depending on direction.

What is the formula to calculate the time for a projectile to reach its maximum height?

The time to reach maximum height is calculated using the formula: t = (v * sin(θ)) / g, where v is the initial speed, θ is the angle of projection, and g is the acceleration due to gravity.

How can you calculate the range of a projectile?

The range of a projectile can be calculated using the formula: Range = (v² * sin(2θ)) / g, where v is the initial speed, θ is the angle of projection, and g is the acceleration due to gravity.

How do you find the final speed of a projectile before it hits the ground?

The final speed (v) before hitting the ground can be calculated using the Pythagorean theorem: v = sqrt(v_x² + v_y²), where v_x is the constant horizontal velocity and v_y is the vertical velocity at the final moment.

Outlines

00:00

🎯 Basic Equations of Motion in Physics

This paragraph introduces an updated version of a previous video on projectile motion, beginning with a review of key kinematic equations. It explains the difference between constant speed and constant acceleration, covering how displacement is calculated under each condition. The paragraph outlines several core formulas, such as how final velocity relates to initial velocity, acceleration, and time. The text also introduces the concept of displacement in both the x and y axes, emphasizing the difference between distance and displacement when direction changes.

05:00

⚖️ The Effect of Gravity on Vertical and Horizontal Velocities

This section explains the impact of gravity on an object’s vertical velocity while horizontal velocity remains constant. Using an example of a ball kicked off the ground with a horizontal velocity of 7 m/s and a vertical velocity of 30 m/s, the text walks through how vertical velocity decreases over time due to gravitational acceleration (approximated at -10 m/s² for simplicity). The relationship between vertical and horizontal velocities over time is explored, with gravity reducing vertical speed until it reaches zero at the top of its trajectory and becoming negative as the ball falls back to the ground.

10:02

🛤️ Types of Projectile Motion and the Role of Angles

This paragraph introduces three common types of projectile motion, beginning with a ball rolling off a cliff. It explains how vertical displacement can be calculated using the equation h = 1/2 a t². The paragraph also discusses how to calculate the horizontal range (r) based on initial horizontal velocity and time. Trigonometric functions (sine and cosine) are used to break down velocity components (v_x and v_y) when an object is launched at an angle, linking them to projectile motion and angles using the Pythagorean theorem.

15:03

🔢 Using the Quadratic Formula in Projectile Problems

This section expands on how to apply the quadratic formula to solve projectile motion problems, particularly when finding the time for a ball to hit the ground. It explains how to calculate the total time of flight by breaking it into the time to reach the highest point (a to b) and the time to descend (b to c). Formulas for range and maximum height are introduced, with a focus on the symmetry of the projectile’s trajectory. The idea that speed at the same height in both ascent and descent is the same is also emphasized.

20:04

🧮 Example Problems on Projectile Motion

This paragraph presents an example of a ball rolling off a 200-meter-high cliff and how to calculate the time it takes to hit the ground. The height of the cliff is derived using the equation h = 1/2 a t², while the horizontal distance (range) is calculated using v_x and time. Another example follows where a ball is dropped from the same height.

Keywords

💡Kinematic equations

Kinematic equations describe the motion of objects under constant acceleration. They help calculate displacement, velocity, and time. In the video, these equations are applied to projectile motion to predict the position and velocity of an object, such as a ball, at various time intervals.

💡Displacement

Displacement refers to the change in an object's position, measured as a vector from the initial to the final position. In the video, displacement is calculated using kinematic equations, particularly for horizontal and vertical components of motion.

💡Acceleration

Acceleration is the rate of change of velocity, typically due to a force like gravity. In projectile motion, the vertical acceleration is constant at -9.8 m/s² (or rounded to -10 in the video), while horizontal acceleration is zero.

💡Projectile motion

Projectile motion describes the trajectory of an object moving under the influence of gravity. The video focuses on how both horizontal and vertical components of motion affect the overall path, such as a ball being kicked or rolling off a cliff.

💡Gravitational acceleration

Gravitational acceleration is the acceleration experienced by objects due to the Earth's gravity, approximately -9.8 m/s². This is a key factor in determining the downward motion of projectiles, as seen in the video’s ball-kicking and cliff-falling examples.

💡Velocity

Velocity is a vector quantity that describes the speed and direction of an object. In the video, both horizontal and vertical velocities are discussed, showing how vertical velocity changes over time due to gravity while horizontal velocity remains constant.

💡Range

Range is the horizontal distance a projectile travels. In the video, it is calculated using the formula Range = velocity × time, showing how far a ball travels horizontally before hitting the ground after being kicked or falling off a cliff.

💡Maximum height

The maximum height is the highest vertical point reached by a projectile during its motion. In the video, the equation for maximum height is derived and used to explain the motion of a ball that is kicked upward and then falls back down.

💡Speed

Speed is the magnitude of velocity and is always positive, while velocity includes both magnitude and direction. The video explains that while the vertical velocity decreases during upward motion, speed decreases and then increases again during the downward fall.

💡Quadratic equation

A quadratic equation is used to solve for time in problems where the projectile’s motion involves vertical displacement. The video shows how to apply the quadratic formula to determine the time it takes for an object to hit the ground when thrown or dropped.

Highlights

Introduction to basic kinematic equations: displacement equals velocity multiplied by time under constant speed.

Displacement under constant acceleration equals the average velocity multiplied by time.

Equations for calculating final velocity: v_final = v_initial + at, and v_final^2 = v_initial^2 + 2a * displacement.

Displacement equation: d = v_initial * t + 1/2 * at^2, where d represents the change in position along the x or y axis.

Explains the difference between distance and displacement when the object moves in one direction versus changing direction.

A projectile's horizontal velocity remains constant, while vertical velocity decreases due to gravity at approximately -10 m/s^2.

At the top of the projectile’s arc, the vertical velocity is zero, but the horizontal velocity remains unchanged.

Gravitational acceleration is the only force acting on a projectile once released, ignoring air resistance.

Velocity versus speed: Speed is the magnitude of velocity and is always positive, while velocity is a vector with both magnitude and direction.

Three types of trajectories: horizontal projection off a cliff, a ball kicked at an angle from the ground, and a ball kicked at an angle from a height.

Key equation for height: h = 1/2 * at^2, derived from vertical motion with gravitational acceleration.

Pythagorean theorem application: v^2 = v_x^2 + v_y^2 to calculate speed just before impact in projectile motion.

Symmetry in projectile motion: The time to reach maximum height is equal to the time to descend to the same horizontal level.

Range of a projectile can be calculated using the formula: range = v^2 * sin(2θ) / g.

The quadratic formula is used to solve for time in complex projectile motion problems when initial speed is given.

Transcripts

this video is going to be an updated

version to an earlier video that i

created on projectile motion but let's

go over some basics the first thing you

need to know are the kinematic equations

at constant speed

d

is equal to vt displacement is equal to

velocity multiplied by the time now

under constant acceleration

displacement is equal to the average

velocity

multiplied by time and the average

velocity

is basically the average of the initial

and the final velocity you add up the

initial and the final and you divide it

by two if the acceleration is constant

now you also have some other equations

v final is equal to v initial plus a t

so the final velocity is equal to the

initial velocity plus the acceleration

multiplied by the time

also

the square of the final velocity is

equal to the square of the initial

velocity

plus two times the product of the

acceleration

and the displacement

and there's another equation

displacement is equal to v initial t

plus

one half

a t squared

now the displacement

is the change in position

it can be the final position

minus the initial position along the x

axis

or it can be along the y axis

so if we replace d

with

y final

minus y initial

we can get another equation that

perhaps you've seen in your physics

course by now

which looks like this the final position

is equal to the initial position

plus v initial t

plus one half

a t squared

now d can represent displacement or

sometimes you can use it to

find distance

distance and displacement are the same

if the object moves in one direction and

doesn't change direction

whenever the object changes direction

displacement and distance are different

so you got to be careful in what you

solve it but if it's moving in one

direction distance and displacement

they're the same in that case

now let's see if we have a ball

and we're gonna

kick it off the ground goes up and then

it's going to go down

gravitational acceleration

is negative 9.8

meters per second squared

for this example we're going to round it

and we're going to use

negative 10 just to keep things simple

so let's say at t equals zero i'm going

to put the time inside the ball

let's say the horizontal velocity v x

we're going to say it's 7

and the vertical velocity v y

we're going to say it's sturdy

now one second later

what do you think

the horizontal and the vertical velocity

will be

what about two seconds later

and then three seconds later

and so forth

so one second later

i should probably put that in a

different color

the horizontal velocity will not change

it's going to remain 7.

the vertical velocity will change

and how much will the vertical velocity

change by

the vertical velocity changes by the

gravitational acceleration acceleration

tells you how much the velocity changes

every second

since the gravitational acceleration

which is the acceleration in the y

direction

it changes about 10 every second

v y is going to decrease by 10 every

second so one second later it's going to

be 20.

two seconds later

the v x is going to be the same but v y

is going to be 10

three seconds later

v y is now zero at the top it's not

going up anymore you only have

horizontal motion

but v x is still seven

then v y becomes negative

the v x will always remain seven

because v x is constant the acceleration

in the x direction is zero

a projectile by definition

is only under the influence of gravity

so if you throw a pen once you release

it from your hand its projectile

a flying bird is not a projectile

its any air but

it has the force

generated by its wings acted on it so

it's not a projectile

projectiles

can only have gravity acting on it air

resistance is ignored

so this is going to be 2 3

4 seconds later

v y

is going to be

negative 10

5 seconds later

negative twenty

six seconds later negative thirty

now here's a question for you

what is the vertical velocity and the

speed five seconds later

the vertical velocity

v y

is negative 20.

the speed

is positive 20.

speed is the absolute value of velocity

velocity can be positive or negative

speed is always positive

speed is the magnitude of velocity

speed is a scalar quantity velocity is a

vector vectors have magnitude and

direction scalar quantities only have

magnitude only

now if you notice because the

acceleration is negative

the velocity

is always decreasing

on the left side it goes from 30 to zero

so it's decreasing on the right side

from zero to negative 30 it's still

decreasing

so the velocity is always decreasing

however the speed

because it's never negative

it's positive on the left side and the

right side

so notice that the speed

is decreasing

on the left side

but on the right side it's increasing

because it goes from 0

to 30.

so whenever the ball

travels upward

the velocity decreases

and the speed decreases but as the ball

begins to fall back down to the ground

or towards the earth

the speed increases and the velocity

continues to decrease

make sure you understand that

now there's three types of trajectories

that you need to be familiar with

so here's the first one

let's say if we have a ball that rolls

off a cliff

and it falls down

h represents the height of the cliff

r is the range

which is the horizontal distance

between the base of the cliff which is

here

and where the ball lands

now because it's moving horizontally

initially the initial speed is v x and

not v

well v and v x are the same

but

the initial speed is v x

at the top whenever it's moving

horizontally v y is equal to zero

as in the case of the other uh problem

so you can easily find the height using

this equation h equals one-half ac

squared

it comes from this equation

now because height is a vertical

displacement

everything in this equation that we use

has to be in the y direction

so we need to use d y vertical

displacement

v y initial

times t plus one half

a y t squared

now d y

is basically the height

at the top v y is zero so this term

disappears and so you get this equation

so make sure you know this equation

h equals one half ac squared is very

useful for this type of trajectory

now if you wish to calculate the range

it's equal to v x t

remember we said that whenever an object

moves with constant speed

d is equal to v t

now v x

is v cosine theta

and v y

is v sine theta

it's based on this triangle here's v

v x

and here's v y and here's the angle

theta

now according to

sohcahtoar perhaps you've seen this if

you've taken trig

s stands for sine

sine theta

is equal to

o and h stands for opposite and

hypotenuse so sine theta is the ratio

between the side that's opposite

to theta which is v y

divided by the sine that's

the hypotenuse which is across the box

and that's v

cosine theta

is equal to the adjacent side vx

divided by the hypotenuse of v

tangent theta

is equal to the opposite side which is v

y divided by the adjacent side v x

so from this equation

if you multiply both sides by v

you can see that v y

is v sine theta

now for the second equation

if you do the same thing v x is v sine

theta

now for the third equation it's useful

to calculate the angle

if tangent theta is v y over v x

then the angle theta

is the inverse tangent of v y divided by

v x

now according to the

pythagorean theorem

there's one more equation that we can

write

since

v x v y and v

are all the three sides in a right

triangle

and according to pythagorean theorem c

squared equals a squared plus b squared

v squared is equal to v x squared plus v

y squared

therefore v

is the square root of v x squared plus

v y squared

if you need to find a final speed just

before it hits the ground

this is the equation that you'll need

now let's talk about the second

trajectory that you'll see

which occurs when a ball is kicked off

from the ground it goes up and then goes

back down

this is going to be the height

or the maximum height of the trajectory

and the horizontal distance is the range

let's call this point a

b

and point c

now if you need to find the time it

takes

to go from point a to point b

it's equal to

v sine theta

divided by g

in another video i show you how you can

derive these equations

now the time it takes to go from

let's say a to c

is simply twice the value

from a to b

due to the symmetry of the graph

so you get this equation

now the maximum height is equal to v

squared

sine squared theta divided by 2g

the range is equal to v squared

sine 2 theta

divided by g

so because the velocity

the ball is kicked at an angle

you can be given v instead of v x or v y

and

the angle relative to the horizontal is

theta

so typically in a problem like this

you'll be given the initial speed any

angle and you could find anything you

need

based on these equations

now the other equations still apply

now it's important to understand that

the speed at part a or point a is the

same as the speed at point c

if you recall

when i went over the trajectory that

looks like this

with all the speeds every second

the speed one second later

was equal to the speed that

occurred at the same height

i think in the last example one second

later the vertical speed was 20 and on

the right side the vertical velocity was

negative 20 which means the speed was

also 20. two seconds later we had 10 and

10.

but as you can see

whenever you have two points at the same

height

the speed will be the same

so if you need to find the final speed

at point c is the same as the initial

speed at point a

just keep in mind at point a the

vertical velocity is positive but at

point c it's negative

but the magnitudes are the same

now the last trajectory

that

you need to be familiar with

looks like this typically you have a

ball

at a cliff or at a building

and it's kicked off at an angle it goes

up and then it goes back down

so all of the equations that you've seen

before

you can apply it

to this trajectory

now in addition to the other equations

one equation that might be useful

is this equation y final equals y

initial

plus

v y initial t

plus one half a t squared or a y t

squared where a y is negative 9.8

y initial

is basically the height of the cliff

and if you want to find the time it

takes to hit the ground that is at point

c

y final replace with zero and then use

the quadratic equation to get the answer

now typically you'll be given the

velocity v

and the angle theta so to find v y

v y is v sine theta so just keep that in

mind

now to use the quadratic formula

make sure you have a zero on one side of

the equation

t

is going to be equal to negative b

plus or minus the square root

b squared minus four 4ac

divided by 2a

now if you don't want to use the

quadratic formula

to find the time it takes to go from a

to c

you can find the time it takes to go

from a to b and add it from b to c

the time it takes to go from a to b we

have that formula

it's going to be 2v sine theta

actually without the two it's v sine

theta

over g

and to find a time it takes

to go from b to c

use this equation you may need to find

the height between

a and b

but if you look at the right side of the

graph between b and c

you can use the fact that h

plus y initial that's the total height

relative to the ground

is equal to one half a t squared

and if you solve for t that's going to

give you the time it takes to go from b

to c

then you add up these two values and you

should get this answer

now another thing that

you may need to know

is you need to calculate the speed just

before it hits the ground

so v x is constant it's going to be the

same at point a b and c

however v y changes

so first you need to find

v y final at point c

using v y initial at point a

once you have v y final and you already

know v x you could find v x by

using this equation

you can calculate the final speed of the

ball just before it hits the ground

using this formula

and to find the angle

use this it's inverse tangent

v y divided by v x

to find the maximum height or the height

between point a and b

you can use this equation it's

v squared

sine squared over

2g

now let's work on some problems

a ball rolls horizontally off a cliff at

20 meters per second

it takes 10 seconds for it to hit the

ground

calculate the height of the cliff and

the horizontal distance traveled by the

ball

now the first thing we need to decide

is which of the three common

trajectories do we

have so the ball

starts off on a cliff

and it rolls horizontally so it's going

to the right and eventually it's going

to fall down

so this is the type of trajectory that

we have in this problem

next

make a list of what you have

we have the speed of the ball as it

leaves a cliff

since it's traveling horizontally

we have the speed vx

so vx is equal to 20 meters per second

now it takes 10 seconds for it to hit

the ground so we have the time

calculate the height of the cliff that's

h

and the horizontal distance

traveled by the ball which is the range

to find the height of the cliff

we could simply use this equation h is

equal to one half a t squared

now a

we can use 9.8

if you use negative 9.8

h is negative but what it really means

is that the vertical displacement is

negative because the ball is going down

but

for all practical purposes we want the

height to be positive so let's plug in

positive 9.8

and t

is 10.

half of 9.8 is 4.9

10 squared is 100.

so 4.9 times 100 is 490.

so that's the height of the cliff

since it takes 10 seconds to hit the

ground

now let's calculate the range

the range is equal to v x t

and we know

v x is v cosine theta but we already

have v x which is 20.

so therefore the range

is going to be 20 meters per second

multiplied by 10 seconds

and you can see that the unit seconds

will cancel

20 times 10

is 200

so the range is 200 meters

and that's it for this problem

a ball rolls off a cliff that is 200

meters high

calculate the time it takes for the ball

to hit the ground

so for this problem

we have a similar trajectory

it rolls off the cliff

and then it hits the ground

and our goal is to find the height

and we know that the height

is one half a t squared

actually we have the height our goal is

to find the time it takes hit the ground

the height is 200.

the acceleration is 9.8

and let's solve for t

so half of 9.8

is 4.9

to isolate t squared we need to divide

both sides by 4.9

200

divided by 4.9

is about

40.8163

[Music]

and that's equal to t squared

so now we need to take the square root

of both sides

to get t

and so t

is going to be

6.389 seconds

and so that's how long it's going to

take

for the ball to hit the ground

a ball is released from rest and drops

straight down starting at a height of

800 meters

how long will it take to hit the ground

now what if the ball was thrown straight

down with initial speed of 30 meters per

second

how long will it take to hit the ground

now

so let's draw a picture

so let's say this is the ground level

and here we have a ball

and we're throwing it down

well for the first one it's going to be

released from rest

and it's going to fall straight down

what is the initial speed if it's

released from rest

now there's no vx component because it's

not moving to the left or to the right

it's falling straight down

so initially

v y

is equal to zero

if it's simply

if it drops down and being released from

rest

now in the other example

v y is not zero

the ball was thrown straight down

so v y

is negative 30. the speed is positive

but

the velocity is negative because it's

going in the negative y direction

in each case

the height is 800 meters

so what can we do to find the time it

takes for it to hit the ground

in the first example you can simply use

the equation h is equal to one half a t

squared

h is eight hundred

the acceleration is nine point eight

and we could solve for t

half of 9.8 is 4.9

800

divided by 4.9

that's

163.265

and that's equal to t squared

so now we can take the square root of

both sides to get t

so t

is 12.78 seconds

now what about part b

how can we find the time it takes for it

to hit the ground if there's an initial

speed

so recall that this equation

comes from this expression

d y

is equal to

v y initial

multiplied by t

plus one half

a y

times t squared

now we need to be careful with the way

we

are going to use this equation

we're going to put the negative signs

in the problem

so we have to be careful

using this equation we didn't have to

worry about the negative signs

now the ball is traveling 800 meters in

the negative y direction so the vertical

displacement is negative 800.

we do have an initial velocity in the y

direction it's negative 30.

so we need to take that into account

and then we have plus one half

a y the acceleration in the y direction

is negative

9.8 so this is going to be

negative 4.9

t squared

now let's move everything from the right

side to the left side

so this is going to be positive 4.9 t

squared on the left side

plus 30t

and the 800 is going to stay on the left

side so it's still negative 800.

as you can see we have a quadratic

equation in standard form to solve for t

we need to use the quadratic formula

so let's make some space first

so here's the quadratic formula t

is equal to negative b plus or minus

square root b squared

minus 4 ac

divided by 2a

so a is 4.9 b is 30 c is negative 800.

so this is going to be negative 30

plus or minus

square root 30 squared is 900

and then this is going to be minus 4

times a which is 4.9

times t squared i mean times c

which is negative 800

divided by 2a

or 2 times 4.9 which is

9.8

so now let's multiply

these three numbers

negative four times four point nine

times negative eight hundred

that's equal to positive

fifteen thousand

six hundred and eighty

and we still have a nine hundred next to

it

so now let's add

those two numbers

if we add 900 it's going to be 16 580

and the square root of that number

is 128.76

now there could be two answers

negative 30 minus 128.76

and plus 128.76

we know time can't be negative so we're

just going to ignore the negative answer

so if we take negative 30

plus 128.76

that's positive 98.76 and divided by 9.8

this will give us

a t value

of

10.08 seconds

which is less than the other answer

which was 12.78

because in the second example

the ball was thrown towards the ground

it's going to take a shorter time to get

to the ground since it was given that

initial speed

and so it makes sense why it would be

less

and we could check the answer

if you plug it into the original

equation

we need to make sure that we get 0. 4.9

times 10.08 squared

plus 30 times

10.08

that's 800.27 minus 800

that's approximately zero

so this answer is correct