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 $=22.7 \mathrm{sec}$