A body is projected vertically upwards with velocity of 98m/s If acceleration due to gravity ceases to act after 5 seconds, distance travelled by the body in the next 5 seconds is

A body is projected vertically up with velocity 98ms^(1) after 2 s if the acceleration due to gravity of earth disappears, the velocity of the body at the end of next 3s is

In the given v-t graph the distance travelled by the body in 5 seconds will be

A body is projected up with a velocity 50 ms^(-1) after one second if acceleration due to gravity disappears then body

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

A body thrown vertically upwards with an initial valocity u reaches maximum height in 6 s. The ratio of the distances traveled by the body in the first second the seventh second is

The distance travelled by a body during last second of its upward journey is d, when the body is projected with certain velocity vertically up. If the velocity of projection is doubled the distance travelled by the body during the last second of its upward journey is

A body falls from rest freely under gravity with an acceleration of 9.8 m//s^(2) . Neglecting air resistance, the distance travelled by the body during the third second of its motion will be:

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))

A particle is projected vertically upwards from ground with an initial speed of 57m/s .If acceleration due to gravity is taken to be equal to 10m/ s^(2) ,then the displacement of the particle in the sixth second of its motion will be?