Introduction to Projectile Motion - Formulas and Equations
Summary
TLDRThis 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(2ฮธ)) / g. Viewers also learn how to avoid common mistakes in calculations and how to apply these equations effectively.
Takeaways
- ๐ The displacement of an object moving at constant speed is equal to velocity multiplied by time.
- ๐ For constant acceleration, the final speed is equal to the initial speed plus acceleration times time.
- ๐ The square of the final speed is equal to the square of the initial speed plus two times acceleration multiplied by displacement.
- ๐ฐ๏ธ Displacement can be found using the average speed (initial speed plus final speed divided by two) multiplied by time.
- ๐ Displacement (D) is equal to initial velocity times time plus half the acceleration times the square of time.
- ๐ In projectile motion, separate calculations for the horizontal (X) and vertical (Y) directions are needed.
- ๐ฐ In horizontal projectile motion, the height (H) can be calculated using H = 1/2 * acceleration * time^2.
- ๐งฎ To find the range of a projectile, use the equation Range = velocity * time in the X direction.
- ๐ฏ The total time of flight for a projectile launched at an angle is 2 * V * sin(Theta) / g.
- ๐ 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
๐ 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.
๐ 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.
๐ 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.
๐ 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.
๐ฏ 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.
๐ 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.
Mindmap
Keywords
๐กProjectile Motion
๐กKinematic Equations
๐กDisplacement
๐กConstant Acceleration
๐กGravitational Acceleration
๐กTrajectory
๐กHorizontal and Vertical Components
๐กRange
๐กTime of Flight
๐กQuadratic Equation
๐กAngle 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
60ยฐ so that's going to be the reference
angle so you can describe it as being
60ยฐ
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
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