Question

Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the same location as Rock X. Both rocks fall for several seconds before landing on the ground directly below the cliff. Frictional forces are considered to be negligible. After Rock Y is released from rest several seconds after Rock X is released from rest, what happens to the separation distance S between the rocks as they fall but before they reach the ground, and why? Take the positive direction to be downward.
A) Sis constant because at the moment Rock Y is released, the only difference between the rocks is their difference in height above the ground.
B) S is constant because the difference in speed between the two rocks stays constant as they fall.
C) S increases because the difference in speed between the two rocks increases as they fall.
D) S increases because at all times Rock X falls with a greater speed than Rock Y

Answer 1

Answer:

D) S increases because at all times Rock X falls with a greater speed than Rock Y

Explanation:

As we know that Stone X is dropped first and then after some time stone Y is dropped

Now we have

$v_x = gt_1$

similarly for Y

$v_y = g(t_1 - \Delta t)$

now the relative acceleration of two stones while they are in air is ZERO

also we know from above equations

$v_x > v_y$

so the separation between two stones will keep on increasing as the speed of X is more than the speed of Y