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Grade: 12
        
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)

Arun
22031 Points
							
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
Khimraj
3008 Points
							
As both are released from rest so initial velocity u = O 
acceleration due to gravity g= 9.8 or say it 10
so for body A
distance s= 16,time t=?
apply s=ut+1/2 gt^2
16=O x t +1/2 x 10 x t^2
16=5 x t^2 say it eqution 1
similarly in 2nd case
u = 0 s= 25 say time as tb for ease
it becomes
25 = 1/2 x 10 x tb^2
25 = 5 tb^2 say it equation 2
now divide equation 1 and 2 we get
16/25 = 5 × t^2 / 5 × tb^2
16/25 = t^2/tb^2
now taking square root on l.h.s 
√16/√25 = t/tb
we know that √16 = √4×4 = 4
√25 =√5×5=5
so ur ans is 4/5
3 months ago
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