A body vertically projected at t=0s and another body is projected at t=2s with same initial velocity 50m/s, then at what time t they meet?

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 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

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 body is projected vertically upwards wth a velocity u=5m//s . After time t another body is projected vertically upward from the same point with a velocity v=3m//s . If they meet in minimum time duration measured from the projection of first body, then t=k/g sec find k (where g is gravitatio acceleration).

A body is dropped from a height h=50 m. Another body is projected vertically up with a speed V=10 m//sec after a time t_0 = 2 sec from the instant of release of the first body. Find the time of their meetings.

A body is projected at time t = 0 from a certain point on a planet’s surface with a certain velocity at a certain angle with the planet’s surface (assumed horizontal). The horizontal and vertical displacements x and y (in meters) respectively vary with time t (in second) as x= 10 sqrt3t , y= 10t -t^2 What is the magnitude and direction of the velocity with which the body is projected ?

A particle is projected from ground with velocity 40sqrt(2)m//s at 45^(@) . At time t=2s

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 horizontally from the top of a tower with a velocity of 20m/s .After what time the vertical component of velocity is four times the horizontal component of velocity (g=10m/s^(2))