Question

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 35 m/s at an angle 32 ∘ above the horizontal. How long was the ball in flight?

Answer 1

Gravitational acceleration on Earth \(=\mathrm{g}=9.81 \mathrm{~m} / \mathrm{s}^{2}\)

Gravitational acceleration on the Moon \(=\mathrm{g}_{\mathrm{M}}\)

The free fall acceleration on the Moon is \(1 / 6\) of that on Earth.

\(g_{M}=\frac{g}{6}\)

\(g_{M}=\frac{9.81}{6}\)

\(g_{M}=1.635 \mathrm{~m} / \mathrm{s}^{2}\)

Initial speed of the ball \(=\mathrm{V}=35 \mathrm{~m} / \mathrm{s}\)

Angle the ball is hit at \(=\theta=32^{\circ}\)

Total time period the ball was in flight \(=\mathrm{T}\)

Time period of flight of a projectile is given by,

\(T=\frac{2 V \operatorname{Sin} \theta}{g_{M}}\)

\(T=\frac{2(35) \operatorname{Sin}(32)}{1.635}\)

\(\mathrm{T}=22.7 \mathrm{sec}\)

Total time period the ball was in flight \(=22.7 \mathrm{sec}\)