A body is projected vertically upwards.At its highest point it explodes into two pieces of masses in the ratio of `2:3` and the lighter piece flies horizontally with a velocity of `6 ms^(-1)`.The time after which the lines joining the point of explosion to the position of particles are perpendicular to each other is

At a certain height a body at rest explodes into two equal fragments with one fragment receiving a horizontal velocity of 10ms^(-1) . The time interval after the explosion for which the velocity vectors of the two fragments become perpendicular to each other is (g=10ms^(-2))

At a certain height a body at rest explodes into two equal fragments with one fragment receiving a horizontal velocity of 10ms^(-1) The time interval after the explosion for which the velocity vectors of the two fragments become perpendicular to each other is (g=10ms^(-2))

At a certain height a body at rest explodes into two equal fragments with one fragment receiving a horizontal velocity of 10ms^−1. The time interval after the explosion for which the velocity vectors of the two fragments become perpendicular to each other is

A body of mass 1 kg is projected vertically upwards wit a speed of 100 m/s. After 6 s it explodes into two parts. One part of mass 400 g comes back with speed 25 m/s. What is the velocity of other part just after explosion?

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 )