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A basketball of mass 0.58 KG and a tennis ball of mass 0.058 KG I dropped from my height of 2.0 M ignoring air resistance what are the kinetic energy and speed of each as they reach the floor?

A basketball of mass 0.58 KG and a tennis ball of mass 0.058 KG I dropped from my-example-1
User Johnkavanagh
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1 Answer

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14 votes

13)

Recall, the law of conservation of energy states that energy is not created nor destroyed but it is converted from one form to another. At the initial positions of the balls above the ground, each has potential energy. As they begin to fall, the potential energy is converted to kinetic energy. As they reach the floor, the energy is fully kinetic

Thus,

potential energy of ball at the top = kinetic energy at the bottom

The formula for calculating potential energy, PE is expressed as

PE = mgh

where

m is the mass of the ball

h is the height above the ground

g is the acceleration due to gravity

The formula for calculating kinetic energy, KE is expressed as

KE = 1/2mv^2

where

m is the mass of the ball

v is the velocity of the ball

Considering the basketball,

m = 0.58

h = 2

g = 10m/s^2

mgh = 1/2mv^2

By substituting these values into the equation, we have

0.58 x 10 x 2 = 1/2 x 0.58 x v^2

11.6 = 0.29v^2

Dividing both sides of the equation by 0.29, we have

11/6/0.29 = 0.29v^2/0.29

v^2 = 40

v = square root of 40

v = 6.32 m/s

The speed of the basketball on hitting the ground is 6.32 m/s

The kinetic energy = 1/2 x 0.58 x 40 = 11.6 J

For the Tennis ball,

m = 0.058

h = 2

0.058 x 10 x 2 = 1/2 x 0.058 x v^2

0.029v^2 = 20

v^2 = 20/0.029 =

User J Whitfield
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