Question

An object is dropped from rest at a height of 196 m and simultaneously another object is dropped from a height of 98 m .what is the difference in their height when both the object fall with equal acceleration after 2 sec of their start

Answer 1

body A is at height 196m

body B is at height 98m

..

As one body is at 196m and another is at 98m the difference in heights between the two will be

= 196m - 98m = 98m

now,

the distance moved by body A in 2s will be

s = ut + (1/2)at²

here,

u = 0

a = 9.8 m/s²

t = 2s

so,

s = 0 + (1/2)x9.8x2²

thus,

s = 19.6m

so, body A will be at height H = 196m - 19.6m = 176.4m above the earth's surface

and

the distance moved by body B in 2s will be

s' = ut + (1/2)at²

here,

u = 0

a = 9.8 m/s²

t = 2s

so,

s' = 0 + (1/2)x9.8x2²

thus,

s' = 19.6m

so, body A will be at height, H' = 98m - 19.6m = 78.4m above the earth's surface

so,

the difference in height after 2s will be

dH = H - H' = 176.4m - 78.4m

thus,

dH = 98m

the height difference would not change as the two bodies move at the same rates.