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A ball is dropped from the top of a cliff. By the time it reaches the ground, all the energy in its gravitational potential energy store has been transferred into its kinetic energy store. If the ball is travelling at 20 m/s when it hits the ground, what height was it dropped from? (Assume that the gravitational field strength is 10 N/kg



Answer :

jcherry99

Answer:

h = 20 meters

Explanation:

Comment

Comment

The total energy (potential energy) it has at the top of the cliff has a formula of PE = mgh. The KE at the top is zero.

The Kinetic Energy at the bottom is 1/2 m v^2 The PE at the bottom is 0.

Givens

g = 10 N/kg

vf = 20 m/s

h = ?

Formula

mgh = 1/2 m v^2   Its the same energy at the top as at the bottom. It just has exchanged  Potential Energy to Kinetic Energy. Cancel the ms

mgh/m = 1/2 m v^2 / m

gh = 1/2 v^2                    Multiply by 2

2gh = v^2                        Put in the given values

2*10 * h = 20^2               Simplify

20h = 400

h = 400/20

h = 20 meters.                        

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