Question

A 20.0 kg rock is on the edge of a 100 m cliff. What is the velocity of the rock just before it hits the grounds when it is pushed off the cliff?

Answer 1

Since the rock is falling we have an uniform accelerated motion. For this kind of motion we know that:

`v_f^2-v_0^2=2a\left(y-y_0\right)`

In this case we have that:

The initial velocity is zero.

The acceleration is 9.8 m/s^2.

The change in position is 100 m.

Plugging this values in the equation we have:

`\begin{gathered} v_f^2-0^2=2\left(9.8\right)\lparen100) \\ v_f^2=1960 \\ v_f=√(1960) \\ v_f=44.27 \end{gathered}`

Therefore, the final velocity is 44.27 m/s