# Physics Bouncing Ball Problem

## Homework Statement

Consider there is a window 5 meters from a person. The window is sized 1.21 meters (4 ft) in length and .914 meter (3 ft) wide at a height of 6 meters from the ground.

A person 1.8288 meters tall (6 ft) wants to bounce a ball (COR: 0.789) on the ground such that the ball goes through the window on the first bounce.

If the person bounces the ball with a speed of 25m/s at an angle of -60° from the horizontal at a point in the middle (between him and the wall), calculate and plot the trajectory of the ball.

Will the ball go through the window? If not, change one of the values such that it does. Then find a RANGE of velocities that will meet this condition, while keeping the COR, Distance, and angle the same. ## Homework Equations

X0(t) = V0cos(theta)t
Y0(t) = V0sin(theta)t - g/2t^2
V = V0(alpha)^n : alpha = the COR, n = the bounce

## The Attempt at a Solution

I formulated an excel sheet to calculate where the ball was at 5 meters along the x direction and discovered that it was too high of a position and would hit above the window. After changing the velocity to 20 m/s it seems to work.

My problem is writing this in a simplistic formula. As well, finding the range of available velocities without having to use an excel spreed sheet for each calculation by "guessing" and trying each available velocity until i find a max and min.

Thanks for any support in this!!! <3

COR

Can you tell me what meant by COR ? The coefficient of restitution (COR), or bounciness of an object is a fractional value representing the ratio of speeds after and before an impact. An object with a COR of 1 collides elastically, while an object with a COR < 1 collides inelastically. For a COR = 0, the object effectively "stops" at the surface with which it collides, not bouncing at all. COR = (relative speed after collision)/(relative speed before collision).

simply multiply this by the velocity after the bounce to get its velocity.

V0*COR = new velocity

or simply V0*COR^n where n is the number of bounces

The coefficient of restitution (COR), or bounciness of an object is a fractional value representing the ratio of speeds after and before an impact. An object with a COR of 1 collides elastically, while an object with a COR < 1 collides inelastically. For a COR = 0, the object effectively "stops" at the surface with which it collides, not bouncing at all. COR = (relative speed after collision)/(relative speed before collision).

simply multiply this by the velocity after the bounce to get its velocity.

V0*COR = new velocity

or simply V0*COR^n where n is the number of bounces

Thanks ! Oonej. But from what i have calculated : The ball would go though the window with above 6.13~6.3 m.

But nevermind ! Just ignore me maybe i got mistakes. Well actually i may have done something incorrectly. can you show me how you did it?

Well actually i may have done something incorrectly. can you show me how you did it?

Same things as you found V1=eVo (normally "e" represents COR) but before you have to find the X1 distance for projected downward motion is equal to 1m.

Then, you have X2 distance for bounced upward motion is 4 m and to get the time ~ 0.4s.

Therefore, using vertical equation for y=V1sin 600t-0.5gt2.

That is my solutions. I must be doing something wrong. How did you solve for the position of X1 at the first bounce? Can you give a more detailed explanation?

I was given a hint by my professor :

Hint: Look at equations for x(t) and y(t). You could eliminate time from the equations and get an equation for y(x)

thanks so much

I must be doing something wrong. How did you solve for the position of X1 at the first bounce? Can you give a more detailed explanation?

I was given a hint by my professor :

Hint: Look at equations for x(t) and y(t). You could eliminate time from the equations and get an equation for y(x)

thanks so much

You so welcome. You just need to apply these two formulae, X1=V0cosθ t and Y1=V0sinθ t + 0.5gt2 to get your X1.(Find up the t from Y1 first)

I have to notice that if you let θ=-600 then your displacement of Y1 also in negative sign.That is why i shown this Y1=V0sinθ t"+"0.5gt2 .

I got .08s for the t where it reaches the ground at x1

I got .08s for the t where it reaches the ground at x1

Yup !

After then you can get X2=5-1=4 m.

Therefore, from X2=V1cosθ t1 => t1=X2/V1cosθ.(t1 is for the range X2 and displacement Y2 of bounced upward motion).

Finally, t1 into this formula(Eliminated the t1):

Y2=V1sinθ t1 - 0.5gt12. To get your answer. Last edited:
Now how do you find the range of velocities to make it through the window?

Will the ball go through the window? If not, change one of the values such that it does. Then find a RANGE of velocities that will meet this condition, while keeping the COR, Distance, and angle the same.

Normally, the Range of X2=V1cosθt1 .So, Range of Velocity is V1cosθ= X2/t1 => VX2=X2/t1. And same things to Range of X1.

Apply this θ=tan-1(e tan 600). Please check it out the answer. Y2 is too low.

Very sorry to you. Last edited:
Now how do you find the range of velocities to make it through the window?

Let say the maximum height for Y2 go through the window is equal to 6 m meanwhile the final vertical velocity Vyf=0 for Y2.

Therefore, apply V2yf=V2yi-2gY2. Because Viy =V1sinθ.So,you can get the V1 after that you can get the V0. I can't understand how he did this. Can anyone reword this so I can comprehend it?

The procedure isn't very clear the way it was explained. Or i'm just slow. lol

I can't understand how he did this. Can anyone reword this so I can comprehend it?

The procedure isn't very clear the way it was explained. Or i'm just slow. lol

This: Here got a mistake please change it ∅=tan-1(e tan θ).

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Can anyone else make sense of this? I greatly appreciate the help Xiaob. But I just dont understand the procedure very well.

I've had my physics teacher look at this and he combined so many formula's that i lost my mind.

Not sure if payment to physics people is allowed on these forums, but I am willing to pay for some help. Consider it tutoring fee's.

Thanks for anyone who helps!

I have discovered that the ball DOES NOT go through the window with the given spec's. But after some brute forcing an excel sheet. I discovered it does go through the window with an angle of -70 degrees at a velocity of 35 m/s

Now i'm stuck at the second part with finding a range.

Ok, more progress...

y2 = x2 tan θ - [ g / ( 2 v1^2 cos^2 θ ) ] * x2^2

using this formula, and changing the degree to 73 (but keeping the velocity at 25 initially which made it through the window. since i was only allowed to change one variable I made it be the angle)

I am getting a few velocities off it seems. minimum calculated out to be 19.1127, but could probably still be a bit slower. and 26.678 for max, but seems it could be a bit faster.

Something I am doing is slightly off. I think it has something to do with the initial x1 value not being calculated correctly.

X2 = 5 - x1, so plugging in 5 wont work, and we don't know what x1 is. Any ideas?

This is one long question

i think we can solve like this:

split initial velocity u into into 2 X,Y components ucosθ and usinθ
during collision and falling X component will experience no change
final velocity in Y, vY can be found out by: vY = e u'
here u' is new value for usinθ after falling 6ft

now for what distance it travels during this first part of motion can be found out by two ways,

I, shorter and using 1 simple formula:
$$y = xtan\theta - \frac{gx^2}{2u^2 cos^2\theta}$$
i always prefer not using this formula

II, general way

use $$y = u_yt + \frac{1}{2}gt^2$$
and find t
now use X = ucosθ t

hence we find uX, uY and distance from point of initial projection

i will continue rest later

Your way doesn't work, when we have 2 unknowns. We can't find T and V at the same time.... or can we?

ok, you have chosen a complicated problem. from what i have read so far your calculations of $$x_1$$ = 1.037 m and $$t_1$$ =0.083 s are correct, though you have rounded too much which is why my results are slightly different.

the situation now gets a little more complicated. the speed in the y-direction increases slightly during this first part of the motion. I calculate the speed in the y-direction right before the bounce to be -22.4636 m/s and so the total speed of the ball just before the bounce is $$v_f$$ = 25.7 m/s.

Also, the angle the ball strikes the ground at will no longer be 60o. the angle I get is now 60.91o.

Here we have to make an assumption; namely that the ball will obey the law of reflection and bounce away at the same angle it came in. this is not unreasonable, but there are plenty of circumstances in which this would not happen. But, let's make that assumption.

ok, so now the ball bounces. it leaves the floor at an angle of 60.91o with a total speed of $$v_2$$ = 20.2773 m/s. thus, $$v_x$$ = 9.86 m/s and $$v_y$$ = 17.72 m/s.

Now, you can calculate how long the ball will take to travel the remaining horizontal distance, and then find the altitude at that time.

In order to find a range of initial velocities, you need to repeat the whole problem, but do not put in a value for $$v_o$$. the result will be an expression for the altitude that depends on $$v_o$$.

hope this helps.

Here we have to make an assumption; namely that the ball will obey the law of reflection and bounce away at the same angle it came in. this is not unreasonable, but there are plenty of circumstances in which this would not happen. But, let's make that assumption.

that is a perfectly wrong assumption

its true only when e=1

Your way doesn't work, when we have 2 unknowns. We can't find T and V at the same time.... or can we?

which part are you talking about?

yeah the reflecting angle is (COR)(THETA)