Question

a projectile is launched from level ground at an initial velocity $v_0$ and angle $\theta$ as shown in the figure. at the instant the projectile is moving horizontally, it hits a wall and then bounces directly back at half the speed it had just before impact. assuming the projectile is in contact with the wall for a negligible time, which of the following expressions represents the total time the projectile is in the air from leaving ground level to reaching ground level again?
(a) $\frac{v_0sin\theta}{g}$
(b) $\frac{3}{2}\frac{v_0sin\theta}{g}$
(c) $2\frac{v_0sin\theta}{g}$
(d) $4\frac{v_0sin\theta}{g}$

Explanation:

Step1: Analyze vertical - motion time to reach maximum height

The initial vertical velocity of the projectile is $v_{0y}=v_{0}\sin\theta$. In vertical - motion, the time $t_1$ taken to reach the maximum height (where the vertical velocity $v_y = 0$) can be found using the kinematic equation $v = v_0+at$. For vertical motion, $v_y = v_{0y}-gt_1$, and when $v_y = 0$, we have $0 = v_{0}\sin\theta - gt_1$. Solving for $t_1$ gives $t_1=\frac{v_{0}\sin\theta}{g}$.

Step2: Analyze vertical - motion time to fall from maximum height

The time taken to fall from the maximum height back to the ground is the same as the time taken to reach the maximum height due to the symmetry of free - fall motion under the influence of gravity. So the time $t_2$ to fall from the maximum height is also $t_2=\frac{v_{0}\sin\theta}{g}$.

Step3: Calculate total time of flight

The total time of flight $T$ of the projectile in the air from leaving the ground and coming back to the ground is the sum of the time to go up and the time to come down. So $T=t_1 + t_2=\frac{v_{0}\sin\theta}{g}+\frac{v_{0}\sin\theta}{g}=2\frac{v_{0}\sin\theta}{g}$. The horizontal motion and the collision with the wall do not affect the vertical - motion time of the projectile.

Answer:

C. $2\frac{v_{0}\sin\theta}{g}$