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A perfectly elastic ball is thrown against a house and bounces back over the head of the thrower. When it leaves the thrower's hand, the ball is $2\text{m}$ above the ground and $4\text{m}$ from the wall and has $V_{0x} = V_{0y} = 10 \text{ m/s}$. How far behind the thrower does the ball hit the ground.assume g = $10 \text{m/s^2}$.

enter image description here I'm stuck in this problem, could anyone give a help? Professor said answer is $2(1 + \sqrt35)$m = $13.8\text{m} $ behind the thrower.

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closed as off-topic by ACuriousMind, HDE 226868, Kyle Kanos, Bernhard, user10851 Aug 17 '15 at 17:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

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Solve the problem with no house, and then reflect it about the position of the wall.

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