Overview of Kinematic Equations for Projectile Motion
Projectile motion problems require understanding and applying basic kinematic equations for constant velocity and constant acceleration. Key equations include:
- Displacement with constant velocity: ( d = vt )
- Final velocity with constant acceleration: ( v = v_0 + at )
- Velocity squared relation: ( v^2 = v_0^2 + 2ad )
- Displacement with average velocity: ( d = \frac{v_0 + v}{2} t )
- Displacement with acceleration: ( d = v_0 t + \frac{1}{2} a t^2 )
Displacement ((d)) refers to the difference between final and initial positions and can be considered separately in the horizontal (x) and vertical (y) directions.
Types of Projectile Trajectories and Relevant Equations
1. Horizontal Launch from a Cliff
- Trajectory: Object launched horizontally from a height (H) and falls to the ground.
- Vertical motion: Use ( H = \frac{1}{2} g t^2 ) to find time to hit the ground.
- Horizontal motion: Range ( R = v_x t ), where ( v_x ) is constant horizontal velocity.
- Vertical velocity at impact: ( v_y = g t )
- Final speed: Combine horizontal and vertical components: ( v = \sqrt{v_x^2 + v_y^2} )
- Impact angle: ( \theta = \tan^{-1}(v_y / v_x) )
2. Projectile Launched at an Angle from Ground Level
- Initial velocity components:
- ( v_x = v \cos \theta )
- ( v_y = v \sin \theta )
- Time to reach max height (point B): ( t = \frac{v \sin \theta}{g} )
- Total flight time (point A to C): ( T = \frac{2 v \sin \theta}{g} )
- Maximum height: ( H = \frac{v^2 \sin^2 \theta}{2g} )
- Range: ( R = \frac{v^2 \sin 2\theta}{g} )
- Final velocity and angle at landing: Same magnitude as launch speed; angle equals launch angle but directed downward.
3. Projectile Launched at an Angle from an Elevated Position
- Initial height: ( H )
- Vertical displacement equation: [ y = H + v_y t - \frac{1}{2} g t^2 ]
- Time to hit ground: Solve quadratic equation for ( t ) when ( y = 0 )
- Alternative time calculation:
- Time to max height: ( t_1 = \frac{v \sin \theta}{g} )
- Time to fall from max height to ground: ( t_2 = \sqrt{\frac{2 y_{max}}{g}} )
- Total time: ( T = t_1 + t_2 )
- Range: ( R = v_x T = v \cos \theta \times T )
- Final velocity components:
- Horizontal velocity remains constant: ( v_x = v \cos \theta )
- Vertical velocity at impact: ( v_y = v \sin \theta - g T )
- Final speed and impact angle: Calculated from velocity components as before.
Additional Notes and Tips
- Always separate horizontal and vertical components when solving projectile problems.
- Horizontal velocity ( v_x ) remains constant (no horizontal acceleration).
- Vertical acceleration is due to gravity ( g = 9.8 \text{ m/s}^2 ).
- Use positive values for velocity components when calculating angles to get reference angles.
- The quadratic formula is essential for solving time in elevated launch problems.
- Understand symmetry in projectile motion to simplify calculations for ground-level launches.
Summary of Key Equations
| Quantity | Equation | Notes | |------------------------|-----------------------------------------------|-------------------------------| | Displacement (constant velocity) | ( d = vt ) | Horizontal or vertical | | Final velocity | ( v = v_0 + at ) | Use separately for x and y | | Velocity squared | ( v^2 = v_0^2 + 2ad ) | Useful for max height | | Displacement with acceleration | ( d = v_0 t + \frac{1}{2} a t^2 ) | Vertical displacement | | Time to max height | ( t = \frac{v \sin \theta}{g} ) | For angled launch | | Total flight time | ( T = \frac{2 v \sin \theta}{g} ) | Symmetrical trajectory | | Maximum height | ( H = \frac{v^2 \sin^2 \theta}{2g} ) | Angled launch | | Range (ground level) | ( R = \frac{v^2 \sin 2\theta}{g} ) | Angled launch | | Range (horizontal launch) | ( R = v_x t ) | Horizontal launch | | Range (elevated launch) | ( R = v_x T ) | Use total time from quadratic |
This comprehensive guide equips you with the essential formulas and methods to confidently solve projectile motion problems across different scenarios. For a deeper understanding of the principles behind these equations, check out Understanding Projectile Motion: A Comprehensive Guide.
To explore the forces and calculations involved in projectile motion, refer to Understanding Projectile Motion: Forces, Velocity, and Calculations.
If you're interested in the energy aspects of projectile motion, see Understanding Kinetic and Gravitational Potential Energy Through Projectile Motion.
For a foundational understanding of motion, you might find Understanding Kinematics: Constant Velocity and Acceleration helpful.
Lastly, to grasp the concepts of linear motion, check out Understanding Linear Motion: Position, Velocity, and Acceleration Explained.
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
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