Question

On the Moon, where air resistance is negligible, an astronaut drops a rock from a cliff and notes that the rock has a speed v after falling from its point of release for a time t. Assuming that the rock does not hit the ground first, how fast will it be moving after it has fallen for a time 4t from its point of release? 2v 4v 8v 16v

Answer 1

Answer:

• 4v

Explanation:

As there are no air resistance, the only force acting on the rock will be the gravitational pull of the moon. Near the surface of the moon, we can take the acceleration from the gravitational pull as constant.

Taking this in consideration, we know that the equation for the speed v at time t will be:

$v(t) = v_0 + a \ t$.

where $v_0$ is the initial speed, and a the acceleration of the object.

Now, as the initial speed is zero, at time t' the speed will be:

$v(t') = a \ t'$.

and at time 4t'

$v(4 t') = a \ 4 \ t'$.

but this is

$v(4 t') = 4 \ (a  \ t')$.

$v(4 t') = 4 \ v(t')$.

So, the answer is 4v.