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 carth. Suppose he hit the ball with a speed of 25 m/s at an angle 30° above the horizontal. a. How long was the ball in flight? b. How far did it travel? c. Ignoring air resistance, how much farther would it travel on the moon than on earth?

Answer 1

15 s

32. 1 m

1/6 times

Step-by-step explanation:

Thinking process:

The acceleration due to gravity =
`(1)/(9)* 9.81 m/s^(2)`

= 1.635 m/s²

Solving for the distance gives

s = ut - 1/2at²

= 33.1 m

Time taken will be:

`t_(2)= 2v\\t_(2) = 2(12.5)(9.81)\\ = 2.55 s`

The distance traveled = vt

=
`21.65 (2.55)\\= 55.2`

Therefore, distance traveled on the surface =

Answer 2

15.34 s

332.1 m

6 times

Step-by-step explanation:

g = (9.81 m/s^2) / 6 = 1.63 m/s^2

Since the ball is affected only by gravity, we can assume this to be a constant acceleration and we can use this equation on the vertical component:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

Y0 will be zero in this case because the ball was launched from the surface

a = -1.63 m/s^2 because gravity points down

Vy0 can be calculated as

Vy0 = V0 * sin(a) = 25 m/s * sin(30) = 12.5 m/s

Y(t) will equal zero in two points, when the ball was initially hit and when it hits the surface again:

0 = Vy0 * t + 1/2 * a * t^2

0 = t * (Vy0 + 1/2 * a * t)

t1 = 0 as expected, this is when the ball was hit

0 = Vy0 + 1/2 * a * t2

1/2 * a * t2 = -Vy0

t2 = -2 * Vy0/a

t2 = -2 * 12.5/(-1.63) = 15.34 s

The difference t2 - t1 = 15.34 s is how long the ball was in flight.

The distance it travelled is calculated with the horizontal speed.

Vx0 = V0 * cos(a) = 25 * cos(30) = 21.65 m/s

Since it is not affected by any horizontal forces, the horizontal speed remains constant. We can use the constant speed equation:

X(t) = X0 + Vx0 * t

X0 can be consifered zero if we deicide to put our origin of coordinates there.

X(15.34) = 21.65 * 15.34 = 332.1 m

X(15.34) - X0 = 332.1 m is how far the ball travelled.

On Earth it would have been affected by a larger gravity.

The time would have been

t2 = -2 * Vy0/a

t2 = -2 * 12.5/(-9.81) = 2.55 s

And travelled

X(15.34) = 21.65 * 2.55 = 55.2 m

It would travel 332.1/55.2 = 6 times further on the Moon than on Earth.