Oblique Projectile Motion

Since both have the same height . We calculate the final velocity of the ball thrown vertically ( Name it B )

Here u = $\beta$

Height , as you know = $u^{2} (\sin\theta)^{2}$ /2g

Therefore velocity of B = $v^{2}$ = $u^{2}$ - 2gh .

Solving this -

$v^{2}$ = $u^{2} (\cos\theta)^{2}$

Meanwhile velocity of A at highest point = Horizontal component of velocity = u $cos\theta$

By conservation of momentum -

mu $\cos\theta$ $\widehat { i }$ + mu $\cos\theta$ $\widehat { j }$ = 2m $\overrightarrow { v }$

$\overrightarrow { v }$ = u/2 $\cos\theta$ $\widehat { i }$ + u/2 $\cos\theta$ $\widehat { j }$

Because of same magnitude , angle with horizontal = 45 $\circ$

All we need to know first, is the velocities of the particles at time of collision.

Let the particle projected at an angle $\alpha$ be $p_1$ , and the particle thrown upward be $p_\2$ .

Right before collision:

$v_{1x} = \beta cos(\alpha)$

$v_{1y} = 0$ (since it is in its highest point).

The only way for the two particles to collide, is to throw $p_2$ later. Suppose $t$ be the time elapsed from $p_2$ is thrown until $p_2$ hits $p_1$ .

$v_{2x} = 0$ (it is projected upward)

$v_{2y} = \beta - gt$

Right before collision, they are of course at the same height. Thus,

$\frac{\beta^2 sin^2 \alpha}{2g} = \beta t - \frac{g}{2} t^2$

Solving for t:

$\frac{g}{2} t^2 - \beta t + \frac{\beta^2 sin^2 \alpha}{2g} = 0$

Using quadratic equation:

$t = \frac{\beta (1 \pm cos \alpha)}{g}$

So:

$v_{2y} = \beta cos \alpha$

Since, it is an elastic collision:

$m\vec{v_1} + m\vec{v_2} = 2m \vec{u}$

$\vec{u} = (\vec{v_1} + \vec{v_2}) / 2$

Thus,

$u_x = \frac{\beta}{2} cos \alpha$

$u_y = \frac{\beta}{2} cos \alpha$

Since,

$tan \gamma = \frac{u_y}{u_x}$

Thus,

$tan \gamma = 1$

$\gamma = \boxed{\frac{\pi}{4}}$