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 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 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 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 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 body is projected upwards. What is its initial velocity at maximum height?

A body is projected upwards with a velocity of 19.6 m/s. Find Greatest height it rises.

The acceleration of a body projected upwards with a certain velocity is

At time t=0 a body is thrown vertically upwards with a velocity "50m/sec" .At time t=2 sec another body is thrown vertically upwards with the same velocity "50m/sec" .the two bodies will meet at time__ sec (a=10m/sec^(2))

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 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.