Two balls are projected simultaneously from the same point, one vertically upwards with a speed 'u' and the other vertically downwards with the same speed 'u'. If they hit the ground with speeds $v_1$ and $v_2$ respectively, then:

A body is fired vertically upward. At half the maximum height, the velocity of the body is 10m/s. The maximum height raised by the body is

A particle starts from the origin of coordinates at time t=0 and moves in the x-y plane with a constant acceleration $\alpha$ in the y-diretion. Its equation of motion is $y = {\rm{\beta }}{x^2}$. Its velocity component in the x-direction is

A bee files a line from a point A to another point B in 4 s with a velocity of $\left| {t - 2} \right|{\rm{m}}{{\rm{s}}^{ - 1}}$. The distance between A and B in metre is

Two tall buildings are 40 m apart. With what speed must a ball be thrown horizontally from a window 145 m above the ground in one building, so that it will enter a window 22.5 m from the ground in the other?

Position of particle during motion is expressed as $\vec r = \left( {5{t^2}\hat i + 3t\,\hat j} \right)m$. The magnitude of acceleration of particle at t = 0, is

A pendulum bob of mass $m$ is released from rest at a height $h$. At what height will its kinetic energy be four times its potential energy?

A ball is dropped on the floor from a height of 10m. It rebounds to a height of 2.5m. If the ball is in contact with the floor for 0.01 s, the average acceleration during contact is nearly [Take ${\rm{g}} = 10{\rm{m}}{{\rm{s}}^{ - 2}}$]

A student is standing at a distance of 50 metres from the bus. As soon as the bus begins its motion with an acceleration of $1\;m{s^{ - 2}}$, the student starts running towards the bus with a uniform velocity u. Assuming the motion to be along a straight road, the minimum value of u, so that the student is able to catch the bus is

The distance travelled by a particle is proportional to the square of time, then the particle travels with

A juggler keeps on moving four balls in the air throws the balls in regular interval of time. When one ball leaves his hand (speed=$20{\rm{m}}{{\rm{s}}^{ - 1}}$), the position of other ball will be (Take g=$10{\rm{m}}{{\rm{s}}^{ - 2}}$)