A moving wall

Relative motion yields a very simple solution. Choose the wall as frame of reference. The ball has Vx = 15m/s, Vy= 5sqrt3 m/s, and its projectile is a parabola with initial angle of 30 degrees (from the ratio Vy/Vx. )

The starting point P is moving horizontally at 10m/s.

When the ball hits the wall, the projectile is reflected. Then it meets P, at 2/3 of the rangebof unreflected parabola, because Vx/10=3/2.

Therefore the wall is at (2/3+1)/2=5/6 of the range of parabola. The time to finish the parabola is 2(5sqrt3/g). Hence the time to hit the wall = (5/6)(10sqrt3/g)=1.47s.

There are two quantities in this problem that can each be calculated in one of two ways.

Total time in air

Let the time from the moment the ball is launched to the moment it contacts the wall be $t_1.$ Similarly, let the time from the moment the ball contacts the wall to when it makes it back to its starting point be $t_2.$

We have the first relation $\boxed{T = t_1 + t_2}.$

Now, the total time spent in the air is given straightforwardly by solving for the second moment at which the ball is touching the ground. We have $v_0\sin\theta\, T - \frac12 gT^2 = 0$ which yields $T = 2v_0\sin\theta/g.$

Distance traveled

Now, the distance traveled by the ball from its launch to its collision with the wall is $d.$ It goes without saying, the distance traveled by the ball from its collision with the wall back to its starting point is also $d.$

Calculated one way we have $v_0\cos\theta\, t_1 = d.$ Calculated another, we have $\left(v_0\cos\theta + 2v_\textrm{wall}\right)\,t_2 = d.$

In the second equation, we use the fact that colliding with the wall reverses the incoming horizontal velocity of the ball and boosts it up by twice the speed of the wall, $v_\textrm{wall}.$

Thus, equating these two expressions for $d,$ we have the second relation

$v_0\cos\theta\, t_1 = \left(v_0\cos\theta + 2v_\textrm{wall}\right)\,t_2,$

or

$\boxed{t_1 = \left(1 + \dfrac{2v_\textrm{wall}}{v_0\cos\theta}\right)t_2}.$

Resolving the system

Solving the two relations for $t_1$ and $t_2,$ we find $t_1 = \frac{v_0\left(2v_\textrm{wall} + v_0\cos\theta\right)\sin\theta}{g\left(v_\textrm{wall} + 2v_0\cos\theta\right)},$ which given the numbers provided, comes out to $t_1 \approx \SI{1.47}{\second}.$