Question

(b) A tennis ball is released from rest at a heighth above the ground. At each bounce 50% of
its kinetic energy is lost to its surroundings. What is the height reached by the ball after its
second bounce?

Answer 1

Answer:

Explanation:

Velocity attained after falling a height h will be v=  

2

 

2gh

​

 

first Bounce-

remaining KE=  

2

mv  

1

2

​  

 

​  

=  

2

1

​  

×  

2

mv  

2

 

​  

 

so velocity in upward direction will be v  

1

​  

=  

2

 

2

​  

 

v

​  

 

so after the time when it return from its top height it will have same velocity as v  

1

​  

 

Now  

for second bounce-

incoming KE is  

2

mv  

1

2

​  

 

​  

 

and after the bounce the remaining KE will be  

2

mv  

2

2

​  

 

​  

=  

2

1

​  

×  

2

mv  

1

2

​  

 

​  

 

so velocity in upward direction will be  v  

2

​  

=  

2

 

2 ​  

 v 1  

= 2 v

so maximum height attained after this bounce will be H=  2g (v  2   )  2

 

=  2g (v/2)  2  

=  8gv2

 

=  8g 2gh

​  

=  4h

​  

 

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