Question

A stone is dropped from rest from the top of a cliff into a pond below. If its initial height is 10 m, what is its speed when it hits the water?is it 14 m/s, 9.9 m/s, or 4.8 m/s

Answer 1

Answer:

14 m/s

Explanation:

The problem can be solved by using the law of conservation of energy. In fact:

- The mechanical energy of the stone at the top of the cliff is just gravitational potential energy (because the stone is at rest, so its kinetic energy is zero), and it is given by

$E=U=mgh$ (1)

where m is the mass of the stone, g=9.8 m/s^2 is the acceleration due to gravity, and h=10 m is the height of the cliff.

- The mechanical energy of the stone when it hits the water is just kinetic energy (because the height of the stone has become zero, so the gravitational potential energy is zero), so it is

$E=K=\frac{1}{2}mv^2$ (2)

where v is the speed of the stone when it hits the water.

Since the mechanical energy is conserved, we can equalize (1) and (2), and solving for v we find:

$mgh=\frac{1}{2}mv^2\\v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(10 m)}=14 m/s$