A body is projected vertically upwards with a speed `v_(0)`. Find the magnitude of time average velocity of the body during its ascent.

A body of mass 2kg is projected vertically upwards with a speed of 3m//s . The maximum gravitational potential energy of the body is :

STATEMENT - 1 : A particle is projected vertically upwards with speed v. Starting from initial point, its average velocity is less than average speed than average speed when particle is returning back to ground. and STATEMENT - 2 : When particle turns back, its displacement is less than distance covered.

A body is projected horizontally with a speed v_(0) find the velocity of the body when it convers equal distance in horizontal and vertical directions.

A body is projected vertically upwards with a speed of sqrt((GM)/R) ( M is mass and R is radius of earth ) . The body will attain a height of

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 upward from the surface of the earth, then the velocity-time graph is:-

From a point on the ground a particle is projected with initial velocity u ,such that its horizontal range is maximum. The magnitude of average velocity during its ascent.

A body is projected vertically upwards with speed u from the top of a tower. It strikes at the ground with speed 3u. What is (a) height of tower and (b) time take by the body to reach at the ground?

A body is projected vertically upward with speed 10 m//s and other at same time with same speed in downward direction from the top of a tower. The magnitude of acceleration of first body w.r.t. second is {take g = 10 m//s^(2) }

A body projected vertically upwards with a given velocity. The ratio of the times taken to cover the first one fourth and the last one fourth of the ascent to the highest point is.