Question

Free Fall On the moon, the acceleration due to gravity is 1.6 \(\mathrm{m} / \mathrm{sec}^{2} .\) (a) If a rock is dropped into a crevasse, how fast will it be going just before it hits bottom 30 sec 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 \(4 \mathrm{m} / \mathrm{sec},\) when will it hit the bottom and how fast will it be going when it does?

Short answer

The answers are (a) the rock will be moving at 48 m/s when it hits the bottom (b) the bottom of the crevasse is 720 m below the point of release and (c) the rock will hit the bottom after 27.5 seconds and at a velocity of 48 m/s.

Step-by-step solution

- Calculate Final Velocity for Part (a)

To solve this, use the formula for velocity under constant acceleration without an initial velocity: Velocity = Acceleration * Time. Given acceleration on moon as 1.6 m/s^2 and time as 30 seconds, Velocity = 1.6 m/s^2 * 30 s = 48 m/s.

- Calculate Distance for Part (b)

To solve this, use formula for distance in case of motion under constant acceleration: Distance = 0.5 * Acceleration * Time^2. Substituting the values given, Distance = 0.5 * 1.6 m/s^2 * (30 s)^2 = 720 m.

- Calculate Time and final Velocity for Part (c)

Firstly, calculate the time it would take for the rock to hit the bottom if thrown at an initial velocity, using the equation of motion: Time = (Final Velocity - Initial Velocity) / Acceleration. However, in this case, we already calculated the final velocity when the rock is dropped (48 m/s), we know the initial velocity is 4 m/s, and the acceleration is 1.6 m/s^2 . So, Time = (48 m/s - 4 m/s) / 1.6 m/s^2 = 27.5 s. Secondly, calculate the final velocity when the rock hits the bottom with the same equation from Step 1. But this time, it has an initial velocity: Final Velocity = Initial Velocity + Acceleration * Time. Substituting the values, Final Velocity = 4 m/s + 1.6 m/s^2 * 27.5 s = 48 m/s. Note that the final velocity is same as that in part (a) as the distance of fall remains the same.