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A body dropped freely from a height H travels 36 percent of total diatmce during the last second. Then H is equal to

A body dropped freely from a height H travels 36 percent of total diatmce during the last second. Then H is equal to

Grade:12

1 Answers

Arun
25750 Points
6 years ago
Let H be the height of the building and t be the time to fall through this height H. 
Then H = ½ g t^2 = 5 t^2.  ---- (1) 
In (t-1) second the distance traveled is 1 - 0.36 H = 0.64 H. 
0.64 H = 5 (t-1)^2.    ------(2) 
(2) / (1) gives 0.64 = (t-1)^2/ t^2 
Or 0.8 = (t-1) / t 
Solving we get t= 5 second. 
Using (1) we get H = 125 m.
 
Regards
Arun (askIITians forum expert)

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