A particle is thrown vertically upward with speed `45 m//s`. Find the distance covered by the particle in the `5^(th)` second of its journey.

A particle is projected vertically upward and moves freely under gravity.If the distance travelled by the particle in the 4^(th) second of its motion is twice of that in the 5^( th ) second,then the speed of projection is ?[g=10m/ s^(2) ]

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

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

A particle is thrown with any velocity vertically upward, the distance travelled by the particle in first second of its decent is

A particle is thrown with any velocity vertically upward, the distance travelled by the particle in first second of its decent is