Question

Two bodies,A(of mass 1kg) and B(of mass 3kg) are dropped from heights of 16 m and 25 m respectively.The ratio of the time taken by them to reach the ground is?

Answer 1

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=t/tb

Answer 2

We know that ut = ½at² ..We are going to use this same equation into he calculation

Here, mass doesn't matter since the acceleration of the gravity acting on every object is same.

For object A,
s = ut + ½at²
16 = 0 + ½(10)(t²)
t² = 16 ÷ 5 = 3.2
t = √3.2 s

For object B,
s = ut + ½at²
25 = 0 + ½(10)t²
t² = 25 ÷ 5 = 5
t = √5 s

Ratio - √3.2 : √5