66.9k views
1 vote
On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The acceleration free-fall on the moon is 1/6 of its value on Earth. Suppose he hit the ball with a speed of 27 m/s at an angle 30 ∘ above the horizontal.

How long was the ball in flight?

User Rakeshdas
by
8.2k points

1 Answer

3 votes

Final answer:

The time a golf ball remains in flight after being hit on the Moon with an initial speed of 27 m/s at a 30-degree angle is 16.875 seconds, considering the Moon's lower gravity and no air resistance.

Step-by-step explanation:

To calculate the time a golf ball remains in flight on the Moon, we will use the projectile motion equations considering the Moon's gravity which is 1/6th of Earth's gravity. The initial speed of the golf ball is given as 27 m/s, and the launch angle is 30 degrees above the horizontal.

We need to find the time of ascent, which is when the ball reaches the highest point, and then double it to find the total time of flight (as the ascent and descent times are equal in a vacuum where there is no air resistance).

The vertical component of the initial velocity is given by:


v_y = v × sin(θ) = 27 × sin(30°) = 27 × 0.5 = 13.5 m/s

Now, using the equation:


t = v_y / g_moon

Where g_moon is the acceleration due to gravity on the moon which is 1.6 m/s².


t = 13.5 m/s / 1.6 m/s² = 8.4375 seconds (time to reach the highest point)

Therefore, the total time the ball was in flight is 8.4375 seconds × 2 = 16.875 seconds.

User Davidmytton
by
8.4k points