Question

1. jacob throws a ball in the air at 18.0 m/s at 30 degrees above the horizontal. just as he throws the ball, bobby drives off in a golf - cart with a constant horizontal speed to catch the ball. a) with what speed must bobby drive away to catch the ball? b) how far away from their initial position does bobby catch the ball?

Explanation:

Step1: Analyze horizontal motion

The horizontal - motion of the ball is a uniform - motion with constant velocity because there is no acceleration in the horizontal direction ($a_x = 0$). The initial horizontal velocity of the ball is given by $v_{0x}=v_0\cos\theta$, where $v_0 = 18.0\ m/s$ and $\theta = 10^{\circ}$. So, $v_{0x}=18\cos10^{\circ}\ m/s\approx18\times0.985 = 17.73\ m/s$.

Step2: Determine the speed of the golf - cart

For Bobby in the golf - cart to catch the ball, the horizontal speed of the golf - cart must be equal to the horizontal speed of the ball. So, the speed at which Bobby drives away is $v = v_{0x}\approx17.7\ m/s$.

Step3: Analyze vertical motion to find time of flight

The vertical - motion of the ball is a uniformly - accelerated motion with $a_y=-g=- 9.8\ m/s^{2}$, $v_{0y}=v_0\sin\theta = 18\sin10^{\circ}\approx18\times0.174 = 3.13\ m/s$. When the ball returns to the same height, the vertical displacement $y - y_0 = 0$. Using the equation $y - y_0=v_{0y}t+\frac{1}{2}a_yt^{2}$, we get $0 = v_{0y}t-\frac{1}{2}gt^{2}=t(v_{0y}-\frac{1}{2}gt)$. One solution is $t = 0$ (corresponds to the initial time). The non - zero solution is $t=\frac{2v_{0y}}{g}=\frac{2\times3.13}{9.8}\ s\approx0.64\ s$.

Step4: Calculate the horizontal displacement

The horizontal displacement of the ball (and also the displacement of the golf - cart) is given by $x = v_{0x}t$. Substituting $v_{0x}\approx17.73\ m/s$ and $t = 0.64\ s$, we get $x=17.73\times0.64\ m\approx11.3\ m$.

Answer:

a) $17.7\ m/s$
b) $11.3\ m$