Question

1. ||| on the apollo 14 mission to the moon, astronaut alan shepard hit a golf ball with a golf club improvised from a tool. the free-fall acceleration on the moon is 1/6 of its value on earth. suppose he hit the ball with a speed of 25 m/s at an angle 30° above the horizontal. a. how long was the ball in flight? b. how far did it travel? c. ignoring air resistance, how much farther would it travel on the moon than on earth?

Explanation:

Part (a)

Step 1: Find vertical component of initial velocity

The initial velocity is \( v_0 = 25\ m/s \) at an angle \( \theta = 30^\circ \). The vertical component \( v_{0y} = v_0\sin\theta \).
\( v_{0y} = 25\sin30^\circ = 25\times\frac{1}{2} = 12.5\ m/s \)

Step 2: Determine acceleration on the moon

Free - fall acceleration on earth \( g = 9.8\ m/s^2 \), so on the moon \( g_m=\frac{g}{6}=\frac{9.8}{6}\approx1.633\ m/s^2 \) (downward, so \( a=-g_m \) in vertical motion)

Step 3: Use vertical motion equation for time of flight

The vertical displacement \( y - y_0 = 0 \) (starts and ends at the same vertical level). The equation is \( y - y_0=v_{0y}t+\frac{1}{2}at^2 \)
Substituting \( y - y_0 = 0 \), \( v_{0y}=12.5\ m/s \), \( a=-g_m=- 1.633\ m/s^2 \)
\( 0 = 12.5t-\frac{1}{2}\times1.633t^2 \)
Factor out \( t \): \( t(12.5 - 0.8165t)=0 \)
Solutions are \( t = 0 \) (initial time) and \( 12.5-0.8165t = 0\Rightarrow t=\frac{12.5}{0.8165}\approx15.3\ s \)

Step 1: Find horizontal component of initial velocity

The horizontal component \( v_{0x}=v_0\cos\theta \)
\( v_{0x}=25\cos30^\circ=25\times\frac{\sqrt{3}}{2}\approx21.65\ m/s \)

Step 2: Use horizontal motion equation (constant velocity)

Horizontal displacement \( x = v_{0x}t \), where \( t \) is the time of flight from part (a) (\( t\approx15.3\ s \))
\( x = 21.65\times15.3\approx331\ m \)

Step 1: Find time of flight on earth

On earth, \( g = 9.8\ m/s^2 \), vertical motion equation \( y - y_0=v_{0y}t+\frac{1}{2}at^2 \), \( y - y_0 = 0 \), \( a=-g=-9.8\ m/s^2 \), \( v_{0y}=12.5\ m/s \)
\( 0 = 12.5t-\frac{1}{2}\times9.8t^2 \)
Factor out \( t \): \( t(12.5 - 4.9t)=0 \)
Solutions \( t = 0 \) and \( t=\frac{12.5}{4.9}\approx2.55\ s \)

Step 2: Find horizontal range on earth

Horizontal range on earth \( x_e=v_{0x}t_e \), \( v_{0x}\approx21.65\ m/s \), \( t_e\approx2.55\ s \)
\( x_e=21.65\times2.55\approx55.2\ m \)

Step 3: Find the difference in ranges

Range on moon \( x_m\approx331\ m \), range on earth \( x_e\approx55.2\ m \)
Difference \( \Delta x=x_m - x_e=331 - 55.2 = 275.8\ m\approx276\ m \)

Answer:

\(\approx15.3\ s\)

Part (b)