motion with constant acceleration question

[CaptainBlood]

A person looks out of the window and sees a tennis ball that appears and disappears twice: first as it rises up past the top of the window and then again as it falls bellow the bottom of the window.  The height of the base of the window to the top of the window is y.  The total time he sees the ball is t. Show that the height (measured relative to the top of the window) that the ball reaches is given by h=1/2g(2y/t - gt/4)^2<br />How do you solve this?  I tried it for an hour and can't figure it out.

Edited March 29, 2011 by CaptainBlood

[swansont]

What is an equation of motion for the ball? i.e. an equation that relates distance speed and time?

[CaptainBlood]

I think the equation to use is y = vt + (1/2)at^2 where v is the initial velocity y is the height, a is acceleration, and t is time, but we're not given initial velocity or the height that the ball reaches. What do I do?

 

I think it has something to do with manipulating the above equation with v^2 = vo^2 + 2 ax where v is the final velocity and vo is the initial velocity, but I have no idea how

[swansont]

You don;t have those values, but you have information about the ball going up and going back down. Using your second equation you can find the speed of the ball at either the top or bottom of the window in terms of the speed at the other point and the size of the window. So you do have information about the initial velocity, if you define that as being when you are first able to see it. And you have the velocity at time t and position y.

 

You should be able to combine equations to get the answer.

 

To break it down: what do you need to get the answer, using the second equation?

[CaptainBlood]

i solved v^2 = vo^2 + 2 ah for vo and got srt(2gh) then put that in to the equation y = vot + (1/2)at^2 for vo, and solved for h, but at the end I get h=1/2g(y/t - gt/2)^2 and not h=1/2g(2y/t - gt/4)^2, so I still don't see how I have to show the 2y/t - gt/4 part. How did he get 2 in the numerator of the first term and 4 instead of 2 in the second. That completely baffles me.

[michel123456]

I suppose if the window was large enough to see the entire path up & down, the resulting equation still stands.

[swansont]

I think the equation to use is y = vt + (1/2)at^2 where v is the initial velocity y is the height, a is acceleration, and t is time, but we're not given initial velocity or the height that the ball reaches. What do I do?

 

I think it has something to do with manipulating the above equation with v^2 = vo^2 + 2 ax where v is the final velocity and vo is the initial velocity, but I have no idea how

 

 

It's a bit of a tricky statement: The total time he sees the ball is t. Which means the time on either the up or down travel is t/2. Your approach was correct. Replace t in your solution with t'/2 and you get the requested solution.

[CaptainBlood]

Great, that does the trick. Thanks man, you're a genius.