Question

Apollo 14 astronaut Alan B. Shepard Jr. used an improvised six-iron to strike two golf balls while on the Fra Mauro region of the moon’s surface, making what some consider the longest golf drive in history. Assume one of the golf balls was struck with initial velocity v0 = 28 m/s at an angle θ = 17° above the horizontal. The gravitational acceleration on the moon’s surface is approximately 1/6 that on the earth’s surface. Use a Cartesian coordinate system with the origin at the ball's initial position.

(a) What horizontal distance, R in meters, did this golf ball travel before returning to the lunar surface?
(b) What is the numeric value of the ratio, f, of the horizontal distance traveled on the moon to the horizontal distance the ball would travel on Earth’s surface under the same conditions?

Answer 1

(a) Rm = 268.4 m

(b) f = 6

Step-by-step explanation:

The horizontal range of a projectile is given by the following formula:

R = V₀² Sin 2θ/g

(a)

For moon:

R = Range on moon = Rm

V₀ = Launch Speed = 28 m/s

θ = Launch Angle = 17°

g = acceleration due to gravity on moon = (9.8 m/s²)/6 = 1.63 m/s²

Therefore,

Rm = (28 m/s)²Sin (2*17°)/(1.63 m/s²)

Rm = 268.4 m

(b)

For Earth:

R = Range on Earth = Re

V₀ = Launch Speed = 28 m/s

θ = Launch Angle = 17°

g = acceleration due to gravity on Earth = 9.8 m/s²

Therefore,

Re = (28 m/s)²Sin (2*17°)/(9.8 m/s²)

Re = 44.7 m

Therefore.

f = Rm/Re = 268.4 m/44.7 m

f = 6