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A body dropped freely has covered (16/25)th of the total distance in the last second. It's total time of fall is?

Answer» Hi.

Here is the answer---


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Let the Total Distance covered by the Body be x and the Total time taken by the body in covering the distance x is y.

Given Conditions,

  Distance covered by the Body in last second = (16/25) of total distance
                                                                         = (16/25) × x
                                                                         = 16x/25

Using the Equation of the Motion for the Free Fall,

  S = ut + (1/2)gt²

Initial Velocity(U) = 0
 [Since the Body starts from the REST at the Certain Height]

 Thus, The Formula will be,

 S = (1/2)gt²

Where, S = Distance
                =  (16/25)x
           
           g = Acceleration due to gravity
               = 10 m/s²
           
            t = time
              = 1 second [ As per as the Question]

Putting these Conditions in the Question,

 We get,
   
  (16/25)x = (1/2) × 10 × (1)²
     16x/25 = 5
      16x = 25 × 5
      16x = 125
         x = 125/16
        x = 7.8125 units.

Total Distance covered by the Body is 125/16 units.

For the Total Time,

 Again using the same equation of the motion,

          S = (1/2)gt²
  125/16 = (1/2) × 10 × t²

     t² =  (125 × 2) ÷ (16 × 10)
\frac{250}{160}     t² =  
\frac{25}{16}     t² = 
     t = √(25/16)
      t  =5/4
      t = 1.25 Seconds.


Thus, the total time taken by the body under free fall is 1.25 Seconds.


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Hope it helps.

Have a nice day.


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