A body is projected veritclaly upwards.The times corresponding to height h while ascending and while descending are `t_(1)` and `t_(2)` respectively.
Then, the velocity of projection will be (take g as acceleration due to gravity)

A body is projected vertically upwards . The times corresponding to height h while ascending and while descending are t_(1)andt_(2) respectively . Then the velocity of projection is (g is acceleration due to gravity)

A body is projected verticallt upwards. If t_(1) and t_(2) be the times at which it is at a height h above the point of projection while ascending and descending respectively, then:

The maximum height reached by projectile is 4 m . The horizontal range is 12 m . The velocity of projection in m s^-1 is (g is acceleration due to gravity)

The maximum height reached by projectile is 4 m . The horizontal range is 12 m . The velocity of projection in m s^-1 is (g is acceleration due to gravity)

A body is vertically upwards. If t_(1) and t_(2) are the times at which it is height h above the point of projection while ascending and descending, respectively, find (a) the velocity of projection and height h, (b) the maximum height reached by the body and (c ) the velocity of the body at height h//2 .

A particle is projected vertically upwards and is at a height h after t_(1) seconds and again after t_(2) seconds then

Two particles are projected vertically upward with the same velocity on two diferent planes with accelerations due to gravities g_(1) and g_(2) respectively. If they fall back to their initial points of projection after lapse of time t_(1) and t_(2) respectively. Then