A 25 kg object is falling towards the earth with a velocity of 8.5 m/s when it is 100 m above the ground. What will be its velocity when it is 20 m above the ground it energy is conserved?

½mv² + mgh = ½mv² + mgh

½ (25.0kg) (8.5 m/s) ² + (25.0 kg) (9.8 m/s²) (100m) = ½ (25.0kg) (v) ² + (25.0 kg) (9.8 m/s²) (20m)

16.2 m/s

40.5 m/s

0.64 m/s

v = 40.5 m/s

Explanation:

Given,

The mass of the object, m = 25 Kg

The velocity of the object at height 100 m, V = 8.5 m/s

The velocity of the object at height 20 m, V = ?

According to the the law of conservation of energy

½ mV² + mgh₁₀₀ = ½mv² + mgh₂₀

Substituting the values in the given equation

½ (25.0kg) (8.5 m/s) ² + (25.0 kg) (9.8 m/s²) (100m) = ½ (25.0kg) (v) ² + (25.0 kg) (9.8 m/s²) (20m)

Solving for v²

v² = 1640.24

v = 40.5 m/s

Hence, the velocity of the object at height 20 m above ground, v = 40.5 m/s