Ceiling height for a game of Catch

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The discussion revolves around calculating the maximum separation distance for two boys playing catch in a hallway with a ceiling height of H, while catching the ball at shoulder height h. The key equation presented is R = 4V(H - h)(vo²/2g - (H - h)), which relates the initial throw velocity to the maximum range. Participants clarify that the range formula is valid but emphasize the need to determine the optimal angle for throwing without hitting the ceiling. The condition H - h > vo²/4g is highlighted for its significance in ensuring the ball reaches the required height. Overall, the conversation focuses on applying physics principles to solve the problem accurately.
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Dear Mates,

I was studying Newton's Laws of Motion by Berkeley's Physics Course, Vol. 1 - Chap. 3, when I came across this problem, about finding the maximum separation of the boys. In my mind, it was simply about finding the "2x", considering x to be the distance between the first boy and the position of the ball at maximum height. Working with the equations of motion for a uniform gravitational field, the task would be resumed in applying the problem's variables to the equation R = Vo²sin2θ/g. But I don't know if this interpretation of mine is wrong of if I've made mistakes in the arithmetics, but I simply can't arrive in the result presented by the own book. And without knowing what I did wrong, I'm unable to give the explanation next requested.

Ceiling height for a game of catch. Two boys "play catch" with a ball in a long hallway. The ceiling height is H, and the ball is thrown and caught at shoulder height, which we call h for each boy, If the boys are capable of throwing the ball with velocity vo, at what maximum separation can they play?

Ans. R = 4V(H - h)[vo²/2g - (H - h)]. Show that if H - h > vo²/4g, R = vo²/g. Explain the physical significance of the condition H - h > vo²/4g.


Thank you for your help,
Cheers,
Luke
 
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The range equation that you have is correct as a starting point because the ball returns to the same height from which it was launched. What do you think you should do next? Show us what you have so far.
 
The range formula is fine as a check, but you do not need to use trig
functions to get the answer provided for (R)^1/2.
 
We know that the height attained by the ball is (v0²sin²ø)/2g<=H-h where H = the highest height attained and h is the height at which the ball is caught by the boys.
Sin ø=√{(H-h)*2g/V0²}
R= V0²/g*[2√{(H-h)*2g/v0²}][√{1-(H-h)*2g/V0²}]...simply to get the result
 
Lukeblackhill said:
applying the problem's variables to the equation R = Vo²sin2θ/g.
But that leaves you with the unknown angle. You need to find the best angle, given that you are not allowed to hit the roof.
 
This is a revived thread from 2 years ago by new user. OP seems to have vanished.
 
kuruman said:
This is a revived thread from 2 years ago by new user. OP seems to have vanished.
Ah, thanks.
 
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