A25 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

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