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
Nor Alo​(Sneha Didi join)

Answer 1

Answer:

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

plsss mark as brainliest i really need it

Answer 2

We know that ut =1/2at^2 ....We

are going to use this same equation into the calculation.

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

For object A,

s= u+ 1/2 at ^2

16= 0 +1/2(10)(t^2)

t^2= 16÷5/3.2

t= root 3.2s

For object B

s=ut +1/2 at ^2

25=0 +1/2(10)+2

t^2= 25÷5=5

t= root 5 s

Ratio: root 3.2:root 5