Weird projectile motion question

Question

The question is as follows:

A ball is thrown from a point $O$ towards a vertical wall in such a way that, after rebounding from the wall, it returns to $O$ without striking the ground. The ball’s initial velocity has magnitude $U$ and is at an angle $θ$ above the horizontal. When the ball strikes the wall, the horizontal component of its velocity is reversed and halved, but the vertical component is unchanged.

(i) Show that $U^2\sin{2\theta}=3gb$, where $b$ is the horizontal distance of the wall from $O$.

(ii) The point $P$ at which the ball strikes the wall is at a height $\frac{2}{9}b$ above the level of $O$. Find $U$ in terms of $b$ and $g$.

(iii) The ball is thrown again from $O$ with the same speed $U$, strikes the wall at the point $Q$, different from $P$ and returns to $O$ without striking the ground. Find, in terms of $b$, the height of $Q$ above the ground.

I found parts (i) and (ii) relatively straight-forward to solve, and I happened to get $U=\sqrt{5gb}$ for part (ii),

My question is: How is at possible that a particle is projected with the same speed from the same point able to follow the same trajectory both ways but hit a different point on the wall? Or am I missing something here?

Answer 1

How is at possible that a particle is projected with the same speed from the same point able to follow the same trajectory both ways but hit a different point on the wall?

Same speed isn't the same thing as same velocity. Two projectiles launched from the same point with the same speed but different angles will follow different trajectories, as they start off with different velocities.

Answer 2

I'm not sure I understand the problem exactly, so do let me know if this doesn't answer your question:

If you've solved the first part, you should be convinced that the particle returns to $O$ when it's projected at a value of $\theta$ that satisfies the following equation: $$\sin{2\theta} = \frac{3 g b}{U^2}.$$ It can be shown that this equation has two roots in the regime $0<\theta<\pi/2$. See, for example, this Math.SE answer to Two roots of arcsin(x) in the range [0,2π]. Essentially, it all boils down to the fact that $\sin{\theta} = \sin{(\pi-\theta)},$ and therefore that $$\sin{2\theta} = \sin(\pi - 2\theta) = \sin\Big(2\left(\frac{\pi}{2} - \theta\right)\Big).$$

In other words, $\theta$ and $\pi/2 - \theta$ are both solutions to the equation, and therefore there are two values of $\theta$ that satisfy the specified relation and consequently two heights that do, too!