A particle is projected vertically upwards and t second after another particle is projected upwards with the same initial velocity. Prove that the particles will meet after a lapse of `((t)/(2)+(u)/(g))` time from the instant of projection of the first particle.

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

A body A is projected upwards with a velocity of 90m//s . The second body B is projected upwards with the same initial velocity but after 4 sec. Both the bodies will meet after

A particle P is projected upwards with 80m//s . One second later another particle Q is projected with initial velocity 70m//s . Before either of the particle srikes the ground (g=10m//s^(2))

A body X is projected upwards with a velocity of 98 ms^(-1) , after 4s, a second body Y is also projected upwards with the same initial velocity . Two bodies will meet after

A particle is projected vertically upward with velocity u from a point A , when it returns to the point of projection .

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 from ground. Which of the following plots best describe the momentum vs height from the ground?