on the moon, the acceleration due to gravity is 1.6m/ sec
1. If instead of being released from rest, the rock is thrown into the crevasse from the same point with a downward velocty of 4m/sec, when will it hit the bottom and how fast will it be going when it does
2. how far below the point of release is the bottom of the crevasse?
To answer these questions, we can use the equations of motion and the principles of kinematics.
1. When will the rock hit the bottom of the crevasse:
We know that the acceleration due to gravity on the moon is 1.6 m/s². Since the rock is thrown vertically downwards with an initial velocity of 4 m/s, we have an initial velocity (u) of -4 m/s (negative because it is in the downward direction) and an acceleration (a) of 1.6 m/s² (positive because gravity is acting downward).
We can use the second equation of motion, which relates distance (s), initial velocity (u), acceleration (a), and time (t):
s = ut + (1/2)at²
Since it is going downward, the displacement (s) is equal to the depth of the crevasse. We can assume the initial position is at the surface of the moon, so the displacement equals the depth of the crevasse.
Let's denote the time taken to reach the bottom as t. Therefore, s = 0 (because the bottom is the final position) and u = -4 m/s, a = 1.6 m/s². Plugging in these values, we get:
0 = (-4)t + (1/2)(1.6)t²
Rearranging the equation, we have:
(1/2)(1.6)t² - 4t = 0
Simplifying further, we have:
0.8t² - 4t = 0
Dividing the equation by 0.8, we get:
t² - 5t = 0
Factorizing, we have:
t(t - 5) = 0
This equation has two roots, t = 0 and t = 5. Since the time cannot be negative, t = 5 seconds.
Therefore, it will take 5 seconds for the rock to hit the bottom of the crevasse.
2. How far below the point of release is the bottom of the crevasse:
To find the depth (s) of the crevasse, we can use the same equation of motion, substituting the values of time (t) and acceleration (a) we obtained in the previous question:
s = (-4) * 5 + (1/2)(1.6)(5²)
Simplifying the equation, we get:
s = -20 + 20
s = 0
Therefore, the bottom of the crevasse is at the same level as the point of release, which means it is 0 meters below the point of release.