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  • An object is dropped from a height h from the ground.  Every time it hits the ground it looses 50% of its kinetic energy.  The total distanc

 

 

  • Option 1) 

    3h

  •  
  • Option 2) 

    \infty

  •  
  • Option 3) 

    \frac{5}{3}h

  •  
  • Option 4) 

    \frac{8}{3}h

Answers (1)

best_answer

Option:(1) is correct.
               When particle starts from 'h' height and hits ground it travels 'h' distance.
               Now its kinetic energy becomes half:
                          kinectic energy just before hitting the ground:                
                                                                                      2gh=1/2mv^2 
                            after hitting it becomes half:
                                                                  1/2(1/2mv^2)
                                    after hitting;
                                                        2gx=1/2(1/2mv^2)
                                                              2gx=1/2(2gh)              \left [ 2gh=1/2mv^2 \right ]
                                                             x=h/2
                                             after hitting it travels h/2 distance upward and h/2 distance downward.
                                        after next hitting it travels h/4 distance upward and h/4 disatnce downward
                                   & this repeats continuously.
So, total distance travelled by particle is:
                                                              h+h/2+h/2+h/4+h/4+h/8+h/8\cdots \infty
                                                             h+h+h/2+h/4+\cdots \infty
                              Sum of GP=
                                                            a/(1-r)                 a=h,r=1/2
                                                             h+ \left \{ h/\left ( 1-1/2 \right )\right \}
                                                             h+2h=3h

Posted by

Rakesh

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