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

The vertical - motion of a projectile is a free - fall motion. The initial vertical velocity is $v_{0y}=v_{0}\sin\theta$. The time it takes for the projectile to reach its maximum height (where the vertical velocity $v_y = 0$) can be found using the kinematic equation $v = v_0+at$. In vertical motion, $a=-g$ (taking up as positive and $g$ is the acceleration due to gravity), and at the maximum height $v_y = 0$. Using $v_y=v_{0y}-gt$, we get $0 = v_{0}\sin\theta−gt_{up}$, so $t_{up}=\frac{v_{0}\sin\theta}{g}$.

Step2: Consider symmetry of vertical motion

The time it takes for the projectile to fall from the maximum height back to the ground is the same as the time it takes to reach the maximum height, $t_{down}=t_{up}=\frac{v_{0}\sin\theta}{g}$. The horizontal - motion and the bounce off the wall do not affect the vertical - motion time. The total time of flight $T$ of the projectile is the sum of the time going up and the time coming down, $T=t_{up} + t_{down}$.

Step3: Calculate total time of flight

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

Answer:

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