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`        Two bodies A (of mass 1kg) and B ( of mass 3 kg) are dropped from height of 16m and 25m respectively.the ratio of the time taken by them to reach the ground is `
3 months ago

## Answers : (2)

```							Let t1 & t2 be time taken to reach the ground then from the formula, h=   1/2 gt^2, For first body,   16  =1/2   gt_(1 )^(   2) For second body,   25 =1/2   gt_2^(   2)  16/25  =(t_1^(  2))/(t_2^(  2) )   ?  t_1/t_2   =  4/5
```
3 months ago
```							As both are released from rest so initial velocity u = O acceleration due to gravity g= 9.8 or say it 10so for body Adistance s= 16,time t=?apply s=ut+1/2 gt^216=O x t +1/2 x 10 x t^216=5 x t^2 say it eqution 1similarly in 2nd caseu = 0 s= 25 say time as tb for easeit becomes25 = 1/2 x 10 x tb^225 = 5 tb^2 say it equation 2now divide equation 1 and 2 we get16/25 = 5 × t^2 / 5 × tb^216/25 = t^2/tb^2now taking square root on l.h.s √16/√25 = t/tbwe know that √16 = √4×4 = 4√25 =√5×5=5so ur ans is 4/5
```
3 months ago
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