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

A body is projected vertically upwards with a velocity of 10 ms^-1 and another body is projected simultaneously from the same point with a velocity of 20 m s^-1 at an angle of pi/6 with the horizontal. The distance between the two bodies after one second from the time of projection is (Acceleration due to gravity is 10 ms^-2 )

A body is projected up with a velocity 50 "ms"^(-1) after one second if accelaration due to gravity disappears then body

A body is projected up with a velocity 50 "ms"^(-1) after one second if accelaration due to gravity disappears then body

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

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 body is projected up with a velocity 50 ms^(-1) after one second if acceleration due to gravity disappears then body