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.

49.5 m/s

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.

Answer it orbits high above earth to gain clearer images of light, it magnifies images relative to the human eye, it records most of the universe as it was in the past, not as it is now;

The answer is 4f======since electrostatic force is given by    we let f be the electrostatic force when  r = 1 m    if we let  denote the electrostatic force when the distance is decreased so that r = 0.5m, then    we established earlier that f = kq₁q₂. therefore    the force will now be 4f.