Question

A ball is dropped from a cliff which is 200 m above the ground below.
1. what is the final velocity of the ball when it hits thee ground below

2. how long does it take for the ball to hit the ground?

Answer 1

From the question, we know that :
h (height of the ball) = 200 m
and also
g (earth's gravity) = 9,8 m/s²

Let :
Em : Mechanical Energy
Ep : Potential Energy
Ek : Kinethical Energy
v : Final Velocitiy when the ball hit the ground
m : mass

Em is always the same wheter the ball is abive the ground or on the ground.

Em = Em

Let the left side is the Em of the ball before it dropped, and the right side when it hit the ground.

Ep + Ek = Ep + Ek
Ep + 0 = 0 + Ek
*because before the ball is dropped it has no velocity, and when it hit the ground it has no potential energy*

m.g.h = $\frac{1}{2}$.m.v²
m.(9,8 m/s²).200 m = $\frac{1}{2}$.m.v²
1960 m²/s² = v²
v = 14√10 m/s

h = $\frac{1}{2}$.g.t²
200 m = $\frac{1}{2}$.(9,8 m/s²).t²
t² ≈ 40,81 s²
t ≈ 6,3 s