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A ball is thrown towards a vertical wall which is a horizontal distance $d$ from the point of projection. The initial speed is $u > 0$ and the angle of projection is $0 < \alpha < \pi/2$. The acceleration due to gravity is $g$ and points downwards.

If the ball hits the wall then the horizontal component of velocity is reversed and the vertical component is left unchanged. The motion can then be considered before the collision with the wall and then after the collision with the wall.

a) Calculate the point at which the projectile strikes the ground as a function of $u$ and $\alpha$

b) Show that in the case that the projectile lands back at the point of projection, the projectile reaches its maximum height at the wall

Any help with how to solve this would be greatly appreciated, I don't really understand which equations etc to use to solve these.

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closed as off-topic by user147263, Bobson Dugnutt, GoodDeeds, Kamil Jarosz, user228113 Mar 11 '16 at 14:36

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HINTS

  1. Determine the vertical and horizontal speeds of the ball in terms of $u$ and $\alpha$.
  2. Now you can do (a) considering just the vertical motion.
  3. (b) follows from symmetry since if it lands where it started, it follows the same trajectory back and forth...
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  • $\begingroup$ @MB You can explicitly compute the maximum height since the projectile itself (in each direction) follows a parabola $S_t = S_0 + v_0 t - gt^2/2$, so the maximum occurs at $t = v_0/g$.... $\endgroup$ – gt6989b Mar 11 '16 at 12:50

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