A particle is thrown upward with speed `20 sqrt(2)` m/s. It strikes the inclined surface as shown in the figure. Collision of particle and inclined surface is perfectly inelastic. What will be maximum height (in m) attained by the particle from the ground `(g = 10 m/s^(2))`

A particle of mass m moving horizontal with v_(0) strikes a smooth wedge of mass M, as shown in figure. After collision, the ball starts moving up the inclined face of the wedge and rises to a height h. The maximum height h attained by the particle is

A particle is projected from the surface of earth with intial speed of 4 km/s. Find the maximum height attained by the particle. Radius of earth = 6400 km and g = 9.8 m//s^(2)

A particle of mass m strikes a disc of radius R and mass m with a speed u as shown in the figure. What is the speed of the particle just after the collision if the collision is perfectly inelastic ?

A particle is projected from the inclined plane at angle 37^(@) with the inclined plane in upward direction with speed 10m//s . The angle of inclined plane with horizontal is 53^(@) . Then the maximum height attained by the particle from the inclined plane will be-

Aparticle is projected from the surface of the earth with an intial speed of 4.0 km//s .Find the maximum height attained by the particle. Radius of earth = 6400km g = 9.8 m//s^(2) .

A ball is projected from the point O with velocity 20 m//s at an angle of 60^(@) with horizontal as shown in figure. At highest point of its trajectory it strikes a smooth plane of inclination 30^(@) at point A. The collision is perfectly inelastic. The maximum height from the ground attained by the ball is (g=10 m//s^(2))

A particle is thrown vertically upward from the ground with some velocity and it strikes the ground again in time 2 s. The maximum height achieved by the particle is : (g=10 m//s^(2))