Question

On the moon, the acceleration due to gravity is 1.6 m/sec2

a) If a rock is dropped into a crevasse, how fast will it be going just before it hits the bottom 30sec. later?
b) How far below the point of release is the bottom of the crevasse?
c) If instead of being released from rest, the rock is thrown into the crevasse from the same point with a downward velocity of 4 m/sec, when will it hit the bottom and how fast will it be going when it does?

Answer 1

The final velocity of a rock dropped on the moon after 30 seconds is 48 m/s, and it would have fallen 720 meters. If the rock is thrown with an initial velocity of 4 m/s, the time and final velocity would differ from the dropped case and require further calculation.

Step-by-step explanation:

The student has asked two related problems based on the acceleration due to gravity on the moon, which is 1.6 m/s².

Part a)

If a rock is dropped into a crevasse on the moon, the final velocity just before it hits the bottom after falling for 30 seconds can be calculated using the formula v = gt, where v is the final velocity, g is the acceleration due to gravity and t is the time. Therefore, v = 1.6 m/s² × 30 s = 48 m/s.

Part b)

The distance fallen can be found using the formula s = (1/2)gt², resulting in s = (1/2) × 1.6 m/s² × (30 s)² = 720 m.

Part c)

When the rock is thrown with an initial downward velocity of 4 m/s, the time to the bottom can be found by solving s = ut + (1/2)gt² where u is the initial velocity. After solving, we find the time and new final velocity as it hits the bottom.