A ball at rest is dropped from a height of 12 m . It loses 25 % of its kinetic energy in striking the ground . Find the height to which it bounces . How do you account for the loss in kinetic energy ?

1 ) A ball at rest is dropped from a height of 12 m .  It loses 25 % of its kinetic energy in striking the ground . Find the height to which it bounces . How do you account for the loss in kinetic energy ?

A ) 5 m ,                                                     B ) 9 m ,

C ) 13 m ,                                                   D ) None of these ,

Correct Answer : -   B ) 9 m ,

Solution : -

So ,  K.E = ( 75 / 100 ) x mgh  =  ( 75 / 100 ) x 12 x mg  =  9 mg .

So , 

P.E  =  9 mg  ,

mgh  =  9 mg  ,

h  =  9 meter   ,   Answer .

Method # 02 : -

Simply we can find 75 % of 12 m   ,  because 25 % is loosed , so we will neglect it .

h  =  12 x ( 75 % )  =  12 x ( 75 / 100 )  =  9 m  ,  Answer .

So it will bounce 9 m after its first strike with ground .

 

2 ) A ball at rest is dropped from a height of 30 m .  It loses 50 % of its kinetic energy in striking the ground . Find the height to which it bounces . How do you account for the loss in kinetic energy ?

A ) 9 m ,                                                     B ) 15 m ,

C ) 21 m ,                                                   D ) None of these ,

Correct Answer : -   B ) 15 m ,

Solution : -

Simply we can find 50 % of 30 m   ,  because 50 % is loosed , so we will neglect it .

h  =  30 x ( 50 % )  =  30 x ( 50 / 100 )  =  15 m  ,  Answer .

So it will bounce 15 m after its first strike with ground .