Question

An object is released from same height exactly after 1 sec another object is released from the same height the distance between the two objects exactly after 2 sec of the release of second object will be

Answer 1

Hey.

As the second object is released after 1s of releasing of first object

so, after 2s of the release of second object time of flight for first object = 1 + 2 = 3 s

time of flight of second object = 2 s

initial velocity for both the objects = 0m/s

a = g = 10 m/s^2 as released from height

As, distance travelled, s = ut + 1/2 at^2

So distance travelled by 1st object, s1

= 0 + 1/2×10×3^2 = 45 m

distance travelled by 2nd object,s2

= 0 + 1/2×10×2^2 = 20 m


Hence, the distance between the two objects exactly after 2 sec of the release of the second object will be

s1 - s2 = 45 - 20 = 25 m

Thanks.