How Is the Rebound Distance of a Ball Calculated After Impact?

[oblong-pea]

Homework Statement

A ball is fired at angle (theta) with velocity (v) from point 0 (the origin) and it follows projectile motion.

It hits a wall at distance (D) from the origin and rebounds.

For this example
theta = 50 deg
V = 50m/s
g = 9.81 m/s^2
D = 200 m

I can plot a graph of the projectile motion, however I'm trying to write an equation to plot the rebound curve of this ball after hitting the wall on a graph with the variables given.

All momentum is conserved, no velocity lost

In this case the ball hits the wall at yIm = 48.41m high from the x axis (0).

I've been stuck on this for a while, so any help would be greatly appreciated.

Relevant Equations

X = horizontal distance
Y = height of ball

For any x distance:
Y = x*tan(theta) - [ (g*x^2) / (2*v^2*cos^2(theta)) ] plots the projectile motion curve.

My attempts involved using suvat equations to determine the rebound distance :
S = 0.5 * (u + v)*t
With u being 50 and v being 0

t being time taken to fall down (Height of impact / gravitational acceleration)
t = 48.41 / 9.81

Plugging the numbers in gives
S = 123.365m

This is where i get stuck.
I tried changing the x values in the projectile equation to between the D and the rebound distance s. It of course plots only that section of the projectile curve. I can't work out how to tell it to plot a rebound from point of impact at yIm to the distance it rebounds by.

 

[PeroK]

How are you modelling the impact with the wall?

Note that momentum is a vector, so i a collision with the wall momentum cannot be conserved. If the collison is elastic, then KE and magnitude of momentum are conserved.

 

[oblong-pea]

How are you modelling the impact with the wall?

Note that momentum is a vector, so i a collision with the wall momentum cannot be conserved. If the collison is elastic, then KE and magnitude of momentum are conserved.

Sorry, I realized i gave a bit of a poor explanation.
Essentially the horizontal rebound velocity is the same as before the collision. So (-) 50m/s.
There are no lossess due to the collison or air resistance and the ground is level.

 

[PeroK]

Sorry, I realized i gave a bit of a poor explanation.
Essentially the horizontal rebound velocity is the same as before the collision. So (-) 50m/s.
There are no lossess due to the collison or air resistance and the ground is level.

The horizontal velocity is not $50 m/s$.

 

[pbuk]

All momentum is conserved, no velocity lost

These statements (assuming they refer to the ball) are not correct. I assume you mean that no kinetic energy is lost in the collision with the wall, i.e. the collision is perfectly elastic.

In this case the ball hits the wall at yIm = 48.41m high from the x-axis (0).

This is correct, so you are on the right track.

For any x distance:
$$ y = x \tan(\theta) - \frac{gx^2}{2u^2\cos^2\theta} $$ plots the projectile motion curve.

This equation is nearly correct (I have corrected it and translated it to $\LaTeX$ so it is easier to read) but you don't seem to have used it. Never mind, it is probably better to stay with first principles.

My attempts involved using suvat equations to determine the rebound distance :
S = 0.5 * (u + v)*t
With u being 50 and v being 0

Where did you get this equation from? If you want to know about the motion of the ball after the collision then you need to know the components of its velocity $( v_x, v_y )$ immediately after the collision. What can we say about these if the collision is perfectly elastic?

Edit: sorry @PeroK, I forgot to hit "Post" on this message when I wrote it before you came in!

 

[oblong-pea]

These statements (assuming they refer to the ball) are not correct. I assume you mean that no kinetic energy is lost in the collision with the wall, i.e. the collision is perfectly elastic.This is correct, so you are on the right track.This equation is nearly correct (I have corrected it and translated it to $\LaTeX$ so it is easier to read) but you don't seem to have used it. Never mind, it is probably better to stay with first principles.Where did you get this equation from? If you want to know about the motion of the ball after the collision then you need to know the components of its velocity $( v_x, v_y )$ immediately after the collision. What can we say about these if the collision is perfectly elastic?

Edit: sorry @PeroK, I forgot to hit "Post" on this message when I wrote it before you came in!

Thanks for your help so far.

Sorry, i understand i missed out the horizontal and vertical components.

So if the launch velocity is 50m/s, then the horizontal velocity is
Uh = 50 x cos(50) = 32.13 m/s

The only information I've been given is that there is no air resistance, the ground is level and the ball rebounds from the wall without loss of velocity. The wall is fixed.

So if there is no loss of velocity on the rebound, would the distance traveled horizontally from the rebound be
S = 0.5 (u+v) t? Or is this an unsuitable suvat to use.
With t still being the height of impact (yIm) divided by g (9.81)?

And if so how would this translate into a equation for y distance to plot as a graph? - Does it rebound at the same angle as the launch angle?

Thanks

 

[PeroK]

And if so how would this translate into a equation for y distance to plot as a graph? - Does it rebound at the same angle as the launch angle?

Thanks

For a perfectly elastic collision: angle of incidence = angle of reflection.