A person is standing on a truck moving with a constant velocity of...

The velocity of the truck is $v_t = 14.7 \text{ m/s}$.

The time taken by the truck to travel $58.8 \text{ m}$ is the time of flight of the ball.

$T = \dfrac{58.8}{14.7} = 4 \text{ s}$

Since the ball returns to the truck, its horizontal velocity as seen from the road must be equal to the velocity of the truck.

$u_x = v_t = 14.7 \text{ m/s}$

The time of flight is given by $T = \dfrac{2u_y}{g}$.

$4 = \dfrac{2u_y}{9.8} \Rightarrow u_y = 19.6 \text{ m/s}$

The speed of projection as seen from the road is $u = \sqrt{u_x^2 + u_y^2}$.

$u = \sqrt{(14.7)^2 + (19.6)^2} = \sqrt{216.09 + 384.16} = \sqrt{600.25} = 24.5 \text{ m/s}$

The angle of projection $\theta$ with the horizontal is given by $\tan \theta = \dfrac{u_y}{u_x}$.

$\tan \theta = \dfrac{19.6}{14.7} = \dfrac{4}{3}$

\theta = 53^{}^{{}^{\circ}}