Question

A 2.4 tonne boulder falls from the top of a 125 m cliff onto the ground below

What speed does it hit the ground with?
(Ignore air resistance.)
Give your answer to the nearest 0.1 m/s

Use g = 9.8 m/s2 for the acceleration due to gravity.

Answer 1

49.5 m/s

Step-by-step explanation:

Let's use the energy approach to solve this problem.

We are given the mass and displacement of the boulder.

The formula for potential energy is PE = mgh.

The formula for kinetic energy is KE = 1/2mv².

If we ignore air resistance, we can use the conservation of energy to solve this problem by setting PE and KE equal to each other. By doing so, we realize that the mass of the boulder does not affect the speed that it hits the ground (final velocity).

• mgh = 1/2mv²

Divide both sides of the equation by m.

• gh = 1/2v²

Let's take the downwards direction as positive. The displacement of the boulder is 125 m and g = 9.8 m/s².

• (9.8)(125) = 1/2v²

Now we can solve for v.

• 1225 = 1/2v²

Multiply both sides by 2.

• 2450 = v²

Take the square root of both sides.

• v = 49.49747468

Rounded to the nearest 0.1 m/s, the final velocity of the boulder is 49.5 m/s.