[" A body is thrown vertically "],[" upwards with "39.2m/sec],[" velocity at the same time "],[" another body is dropped from "],[" 100m height.After how long "],[" the bodies acquires same "],[" velocity."]

[" A body is thrown up with a "],[" velocity "40ms^(-1)" .At same time "],[" another body is dropped from "],[" a height "40m" .Their relative "],[" acceleration after "1.3],[" seconds is "]

A body is throw vertically upward with a velocity of 9.8 m/s. Calculate the maximum height attained by the body.

A body A is thrown vertically upwards with such a velocity that it reaches a maximum height of h. Simultaneously, another body B is dropped from height h. It strikes the ground and does not rebound. The velocity of A relative to B versus time graph is best represented by (upward direction is positive).

A body is thrown up with a velocity 40 ms ^(-1). At same time another body is dropped from a height 40 m. Their relative acceleration after 1.3 seconds is

A body is thrown vertically upward with a velocity of 9.8 m/s. Calculate the maximum height attained by the body. (g=9.8m//s^(2))

For a body thrown vertically up with an initial velocity u ,the maximum height reached by body from the ground is

A body A is thrown vertically upwards with initial velocity v_(1) . Another body B is dropped from a height h. Find how the distance x between the bodies depends on the time t if the bodies begin to move simultaneously.

At time t=0 a body is thrown vertically upwards with a velocity "50m/sec" .At time t=2 sec another body is thrown vertically upwards with the same velocity "50m/sec" .the two bodies will meet at time__ sec (a=10m/sec^(2))

A body is thrown vertically upwards and rises to a height of 10m. The velocity with which the body was thrown is (g - 9.8m/( s^(2) )

A body is thrown vertically upwards and takes 5 seconds to reach maximum height. The distance travelled by the body will be same in :-