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

Question

Two bodies, A (of mass 1 kg) and B (of mass 3 kg) are dropped from heights of 16 m and 25 m, respectively. The ratio of the time taken by them to reach the ground is:- (A) 5/4 (B) 12/5 (C) 5/12 (D) 4/5

Tardigrade Admin · Accepted Answer

from equation of motion we know that when a body is dropped h =(1/2) gt 2 So ( t 1/ t 2)=√( h 1/12) =√(16/25)=(4/5)

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

5/4

12/5

5/12

4/5

from equation of motion we know that when a body is dropped
$h = \frac{1}{2} g t^{2}$
So $\frac{t _{1}}{t _{2}} = \frac{h _{1}}{1 _{2}}$
$= \frac{16}{25} = \frac{4}{5}$

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

5/4

12/5

5/12

4/5

from equation of motion we know that when a body is dropped h=21​gt2 So t2​t1​​=12​h1​​​ =2516​​=54​

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

from equation of motion we know that when a body is dropped
$h = \frac{1}{2} g t^{2}$
So $\frac{t _{1}}{t _{2}} = \frac{h _{1}}{1 _{2}}$
$= \frac{16}{25} = \frac{4}{5}$