mass ‘m’ is projected vertically upwards with a speed ‘u’ and at the same instance another particle of mass ‘m’ is released from height ‘h’ vertically above the first particle. The two particles suffer a perfectly inelastic collision, and again reach a maximum height ‘h’ as one combined particle. What is the value of u?

A particle is projected vertically upwards in vacuum with a speed u .

A particle is projected vertically upward in vacum with a speed 40m//s then velocity of particle when it reaches at maximum height 2s before, is (Take g=10m//s )

A particlele is projected vertically upwards from the ground with a speed V and a speed v and a second particle is projected at the same instant from a height h directly above the first particel with the same speed v at an angle of projection theta with the horizontal in upwards direction. The time when the distance between them is minimum is

A particle is projected vertically upwards and it reaches the maximum height H in time T seconds. The height of the particle at any time t will be-

A particle is projected vertically upwards and reaches the maximum height H in time T. The height of the partlcle at any time t (lt T) will be

If two particles of masses m_(1) and m_(2) are projected vertically upwards with speed v_(1) and v_(2) , then the acceleration of the centre of mass of the system is

A particle of mass m moving with speed u collides perfectly inelastically with another particle of mass 3 m at rest. Loss of KE of system in the collision is

A particle is projected vertically upward from the ground at time t = 0 and reaches a height h at t = T . Show that the greater height of the particle is (g T^(2) + 2h)^(2)//8 gT^(2)

An object is projected vertically upward from the surface of the earth of mass M with a velocity such that the maximum height reached is eight times the radius R of the earth. Calculate: (i) the initial speed of projection (ii) the speed at half the maximum height.