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 hits the ball with a speed of 25 m/s at an angle of 28° above the horizontal.
(a) How long was the ball in flight?
(b) How far did it travel?
(c) How much farther would it have traveled on the moon than on earth?

Answer 1

Vo = 25m/s @ 28 Deg.
Xo = 25*cos28 = 22.1 m/s.
Yo = 25*sin28 = 11.7 m/s.

b. Dx = Vo^2*sin(2A)/g.
Dx = 25*sin(56)/1.63 = 317.9 m.

a. Dx = Xo*T = 317.9 m.
22.1 * T = 317.9.
T = 317.9 / 22.1 = 14.38 s. = Time in
flight.

c. D'x = 317.9 - (317.9/6) = 265 m.

Answer 2

whoever already answered this question was very, very incorrect. your formulas may be, but your own calculations don't even come out to the numbers you have.

Answer 3

for b, you should have used 25Cos28*14.38.

Answer 4

g_moon=9.8 / 6=1.633333333333 m/s²
t/2=v sin Θ/g
=25 x 11.736789/1.633333333333
=7.18579
t =14.3715785 seconds in the air
Range=v² sin 2Θ/g
=625 x 0.507574024
=317.233765m as the range of ball on the moon

Range_earth=625 sin 56 / 9.8
=52.8723m as the distance on earth.........