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Firstly sorry if this has been asked before. I'm trying to work out a simple graphical equation that'll give me the x of the bounce point when y=0 (as it's hit the ground).

I'm only after the first bounce and the peak of the second bounce (so essentially it's like throwing a ball at a wall, getting it's first bounce and the point at which it hits the wall).

Gravity and the weight of the ball will always be constant, there will be no air resistance and the only variables will be the height from which the ball is thrown and the speed it's thrown at.

Can you help me out?

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    Let me make sure I understand. We will throw a ball with an initial height and velocity velocity (note that I mean both speed and direction). We assume no obstructions other than a perfectly flat ground, and any and all collisions will be perfectly elastic (the initial energy of the ball is conserved). Is that correct?2011-07-18
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    @Ahmed I do not fully understand your question. I am making a number of assumptions here; a better answer can be provided if you make the question clearer. I am assuming that you are throwing a ball *horizontally* with a velocity $u$ from a height $h$. (Acceleration to gravity is $g \approx 9.81$.) In this case, the $x$ coordinate of the first bounce is given by the formula $u \sqrt{\frac{2h}{g}}$. I am also assuming that the collision with the ground is *perfectly elastic* (i.e., it rebounds with the same speed). In this case, the ball will reach a maximum height of $h$ after the first bounce2011-07-18
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    @Srivatsan: Why don't you try to answer or ask one or two questions to get the required 50 rep so that you can comment properly and not bother the moderators with the "comments posted as answers"?2011-07-18
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    (this is Srivatsan's comment continued): is given by the formula $u \sqrt{\frac{2h}{g}}$. I am also assuming that the collision with the ground is perfectly elastic (i.e., it rebounds with the same speed). In this case, the ball will reach a maximum height of $h$ after the first bounce.2011-07-18
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    Sorry for not explaining it correctly. I should have mentioned that I want some sort of gravity constant in there so that when it rebounds, it rebounds at a degraded height. The path of the ball will be on a 2D axis, so imagine the path across your screen going from left (where we throw it from), bouncing about 3/4 of the way and touching the right side of your screen.2011-07-19

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