Question

a golfer strikes a golf ball that is initially on the ground. the ball leaves the ground with initial speed $v_0$ at an angle $\theta$ to the horizontal and travels toward a vertical wall. the ball collides with the wall at the instant it reaches what would have been its highest point if the wall wasnt there. the ball rebounds from the wall with the same speed it had when it struck the wall, and then lands at the same point from which it was struck. which of the following is a correct expression for the horizontal distance between the balls initial position and the wall?
a $\frac{v_0^2sin\thetacos\theta}{2g}$
b $\frac{v_0^2sin\thetacos\theta}{g}$
c $\frac{v_0^2sin^2\theta}{2g}$
d $\frac{v_0^2sin^2\theta}{g}$

Explanation:

Step1: Analyze vertical - motion time

The time $t$ it takes for a projectile to reach its maximum height is given by the kinematic equation $v = v_0+at$ for vertical motion. In vertical motion, the initial vertical velocity is $v_{0y}=v_0\sin\theta$, the acceleration $a=-g$, and at the maximum - height, the vertical velocity $v_y = 0$. Using $v_y=v_{0y}-gt$, we get $0 = v_0\sin\theta−gt$. Solving for $t$, we have $t=\frac{v_0\sin\theta}{g}$.

Step2: Analyze horizontal - motion distance

The horizontal motion of a projectile is a uniform - motion with constant horizontal velocity $v_{0x}=v_0\cos\theta$. The horizontal distance $x$ is given by the formula $x = v_{0x}t$. Substituting $t=\frac{v_0\sin\theta}{g}$ and $v_{0x}=v_0\cos\theta$ into the horizontal - distance formula, we get $x=(v_0\cos\theta)\times\frac{v_0\sin\theta}{g}=\frac{v_0^{2}\sin\theta\cos\theta}{g}$.

Answer:

B. $\frac{v_0^{2}\sin\theta\cos\theta}{g}$