A stone is projected from a horizontal plane. It attains maximum height $H$ and strikes a stationary smooth wall and falls on the ground vertically below the maximum height. Assume the collision to be elastic, the height of the point on the wall where the ball will strike is: \[ \begin{array}{ll} (1)

Key ideas you must know

1. Projectile motion basics
   • Horizontal motion: $x = u\cos\theta\,t$ (uniform)
   • Vertical motion: $y = u\sin\theta\,t - \tfrac12 g t^2$ (accelerated)

2. Maximum height $H$
   The ball’s vertical velocity becomes zero at time
   $t_p = \frac{u\sin\theta}{g}$
   giving
   $H = \frac{u^2\sin^2\theta}{2g}$

3. Elastic collision with a smooth vertical wall
   • Only the horizontal component of velocity reverses ($u_x \to -u_x$).
   • Vertical component is unchanged.
   Therefore speed is conserved.

4. Condition given
   After the bounce the stone lands directly below its maximum-height point. That means:
   $x_{\text{land}} = x_{\text{peak}}$

5. Unknowns
   • $t_1$ = time from launch to wall-hit
   • $t'$ = time from wall-hit to landing
   • $y_1$ = height where it hits the wall (what we need)

6. Equations you set up
   • Horizontal positions
   $x_{\text{peak}}\;=\;u\cos\theta\,t_p$
   $x_{\text{wall}}\;=\;u\cos\theta\,t_1$
   After bounce: $x = x_{\text{wall}} - u\cos\theta\,t'$
   Setting $x_{\text{land}} = x_{\text{peak}}$ gives
   t_1 - t' = t_p \tag{1}

   • Vertical motion after bounce must return to the ground:
   y_1 + (u\sin\theta - g t_1)\,t' - \tfrac12 g t'^2 = 0 \tag{2}

   • Algebraic manipulation of (2) surprisingly yields
   t_1 + t' = 2t_p \tag{3}

   Solve (1) and (3):
   $t_1 = \tfrac32 t_p,\quad t' = \tfrac12 t_p$

7. Height of wall-hit
   Substitute $t_1$ into the vertical equation for $y$:
   $y_1 = u\sin\theta\,t_1 - \tfrac12 g t_1^2$
   $\;\;\;\; = u\sin\theta\left(\tfrac32 t_p\right) - \tfrac12 g\left(\tfrac32 t_p\right)^2$
   $\;\;\;\; = \tfrac32 g t_p^2 - \tfrac98 g t_p^2 = \tfrac38 g t_p^2$

   Now compare with $H = \tfrac12 g t_p^2$:

   $y_1 = \frac{3}{8}g t_p^2 = \frac{3}{4}\Bigl(\tfrac12 g t_p^2\Bigr)=\frac{3}{4}H$

Hence the ball meets the wall at $\boxed{\tfrac34 H}$.

Imagine this

1. You throw a ball from the ground so that it goes up in a nice curved path (a ‘rainbow’ shape) and would normally land somewhere far away.
2. Half-way through its flight the ball reaches its highest point. We call that height $H$.
3. Now place a smooth, perfectly bouncy wall somewhere in front of you. When the ball hits the wall, it bounces sideways without losing any speed (elastic collision).
4. After the bounce the ball comes straight back and finally lands exactly under the spot where it had been highest.

Because the bounce is perfect, only the sideways (horizontal) speed changes direction; the up-down (vertical) speed stays the same. By stitching together the timing before and after the bounce, we can work out how high on the wall it must have hit. The maths shows that the hit-point is at three-quarters of the maximum height, i.e.

$\text{height on wall} = \frac{3}{4}H$

Step-by-step Mathematical Solution

1. Let the initial speed and angle be $u$ and $\theta$.
   Horizontal velocity: $u_x = u\cos\theta$
   Vertical velocity: $u_y = u\sin\theta$

2. Time to maximum height
   $t_p = \frac{u\sin\theta}{g}$

3. Maximum height
   $H = \frac{u^2\sin^2\theta}{2g}$

4. Unknowns
   $t_1$ = time to reach the wall,
   $t'$ = time from wall to ground,
   $y_1$ = required wall-hit height.

5. Horizontal position condition
   $x_{\text{peak}} = u\cos\theta\,t_p$
   $x_{\text{wall}} = u\cos\theta\,t_1$
   After bounce: $x_{\text{land}} = x_{\text{wall}} - u\cos\theta\,t'$
   Given $x_{\text{land}} = x_{\text{peak}}$:

   $u\cos\theta(t_1 - t') = u\cos\theta\,t_p \;\Longrightarrow\; t_1 - t' = t_p \qquad (1)$

6. Vertical motion after collision to hit the ground
   Post-collision vertical velocity remains $u\sin\theta - g t_1$.
   Setting $y=0$ at landing:

   $y_1 + (u\sin\theta - g t_1)t' - \tfrac12 g t'^2 = 0 \qquad (2)$

   where
   $y_1 = u\sin\theta\,t_1 - \tfrac12 g t_1^2$

   Insert $y_1$ into (2) and simplify:

   $u\sin\theta(t_1 + t') - \tfrac12 g(t_1 + t')^2 = 0 \;\Longrightarrow\; u\sin\theta = \tfrac12 g (t_1 + t')$

   Hence
   $t_1 + t' = \frac{2u\sin\theta}{g} = 2t_p \qquad (3)$

7. Solve (1) and (3)
   Adding: $2t_1 = 3t_p \;\Rightarrow\; t_1 = \tfrac32 t_p$
   Subtracting: $2t' = t_p \;\Rightarrow\; t' = \tfrac12 t_p$

8. Height at the wall
   $y_1 = u\sin\theta\left(\tfrac32 t_p\right) - \tfrac12 g\left(\tfrac32 t_p\right)^2$

   Replace $u\sin\theta$ by $g t_p$:

   $y_1 = \tfrac32 g t_p^2 - \tfrac98 g t_p^2 = \tfrac38 g t_p^2$

   Compare with $H = \tfrac12 g t_p^2$:

   $y_1 = \frac{3}{4}H$

9. Final answer
   [ \boxed{\displaystyle \text{Height on wall} = \frac{3H}{4}} ]