A particle starts from rest and moves with a varying acceleration as shown in the graph. If the particle's velocity at $t=4$ s is $v$, its velocity at $t=8$ s will be:

A balloon is at a height of 81 m and is ascending upwards with a velocity of 12 m/s . A body of 2kg weight is dropped from it. If $g = 10\;m/{s^2}$, the body will reach the surface of earth in

A particle moves such that its displacement $x$ varies with time $t$ as $x = At^2 + Bt^3$. The graph of velocity versus time will be:

A body of mass 3 kg falls from the multi-storeyed building 100m high and buries itself 2m deep in the sand. The time of penetration will be

Two balls A and B are thrown simultaneously from the top of a tower. A is thrown vertically up with a speed of $4\;{\rm{m}}{{\rm{s}}^{ - 1}}$. B is thrown vertically down with a speed of $4\;{\rm{m}}{{\rm{s}}^{ - 1}}$. The ball A and B hit the ground with speed ${v_A}$ and ${v_B}$ respectively. Then,

Two balls of same size but the density of one is greater than that of the other are dropped from the same height, then which ball will reach the earth first (air resistance is negligible)?

Two balls A and B of same masses are thrown from the top of the building. A, thrown upward with velocity V and B, thrown downward with velocity V, then

The slope of a distance-time graph represents:

A ball is thrown vertically upwards from the top of a tower at $4.9\;m{s^{ - 1}}$. It strikes the pond near the base of the tower after 3 seconds. The height of the tower is

Two balls of same size but the density of one is greater than that of the other are dropped from the same height, then which ball will reach the earth first (air resistance is negligible)?

From the top of a tower of height 50m, a ball is thrown vertically upwards with a certain velocity. It hits the ground 10 s after it is thrown up. How much time does it take to cover a distance AB where A and B are two points 20m and 40m below the edge of the tower? (${\rm{g}} = 10{\rm{m}}{{\rm{s}}^{ - 2}}$)