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

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.