A body is projected vertically up. The distance that it covers in the last one second of its ascent is

If a ball is thrown vertically upwards with speed u ,the distance covered during the last T seconds of its ascent is

If a ball is thrown vertically upwards with speed u , the distance covered during the last t second of its ascent is

A body is projected vertically upward with speed 40m/s. The distance travelled by body in the last second of upward journey is [take g=9.8m/s^(2) and neglect eggect of air rresistance]:-

" A body is projected vertically up and the total distance travelled by it is 'S' Then total time of flight will be "

STATEMENT - 1 : Two bodies projected vertically upward with different initial speeds cover same distances in last second of their upward journey. and STATEMENT - 2 : For uniformly accelerated motion, distance travelled in a time interval before coming to rest does not depend on initial speed.

With what velocity a ball be projected vertically so that the distance covered by it in 5^(th) second is twice the distance it covers in its 6^(th) second (g=10 m//s^(2))

A body is projected vertically up with a speed of 20ms^(-1) .The distance travelled by it in "3" seconds it (g=10ms^(-2))

The distance travelled by a body during last second of its total flight is d when the body is projected vertically up with certain velocity. If the velocity of projection is doubled, the distance travelled by the body during last second of its total flight is (A) 2d (B) d (C) 2d + g/2 (D) 2d –g/2