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Readme [11.4K]
2 years ago
7

You drop a 0.375 kg ball from a height of 1.37 m. It hits the ground and bounces up again to a height of 0.67 m. How much energy

did it lose in the bounce?
Physics
1 answer:
Radda [10]2 years ago
6 0

2.57 joule energy lose in the bounce .

<u>Explanation</u>:

when ball is the height of 1.37 m from the ground  it has some gravitational potential energy with respect to hits the ground  

Formula for gravitational potential energy given by  

Potential Energy = mgh

Where ,

m = mass  

g = acceleration due to gravity  

h = height

Potential energy when ball hits the ground

m= 0.375 kg

h = 1.37 m

g = 9.8 m/s²

Potential Energy = 0.375\times9.8\times1.37

Potential Energy = 5.03 joule

Potential energy when ball bounces up again

h= 0.67 m

Potential Energy = 0.375\times0.67\times9.8

Potential Energy = 2.46 joule

Energy loss = 5.03 - 2.46 = 2.57 joule

2.57 joule energy lose in the bounce

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Answer:

The height of Sears Tower is 1448.5 feet.

Explanation:

<h3>We apply the free fall formula to the ball: </h3><h3>y=v_{o} *t+\frac{1}{2} *g*t^{2}</h3><h3>y: The vertical distance the ball moves at time t  </h3><h3>v_{o}i: Initial speed </h3><h3>g=Gravity acceleration=9.8*(\frac{\frac{1ft}{0.305m} }{s^{2} } )</h3>

Known information

We know that the vertical distance (y) that the ball moves in 9,5s  is equal to height of Sears Tower (h).  

Too we know that the ball is released from rest, then,v_{0}=0

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We replace  in the equation 1 the data following;

y=h

v_{o} =0

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h=0*9.5+\frac{1}{2} *32.1*9.5^{2}

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b) v = 15.8 m/s

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