$A$ balloon is rising vertically up with a velocity of $29 \,ms^{-1}$. $A$ stone is dropped from it and it reaches the ground in $10 \,s$. The height of the balloon when the stone was dropped from it is...........$m$ $(g = 9.8 \,ms^{-2})$

$A$ balloon rises up with a constant net acceleration of $10\,m/s^2$. After $2\,s$,a particle drops from the balloon. After a further $2\,s$,match the following columns. (Take $g=10\,m/s^2$)

Column $I$	Column $II$
$(A)$ Height of particle from ground	$(p)$ $0$
$(B)$ Speed of particle	$(q)$ $10$ $SI$ unit
$(C)$ Displacement of particle	$(r)$ $40$ $SI$ unit
$(D)$ Acceleration of particle	$(s)$ $20$ $SI$ unit

$A$ ball is dropped from a height $h$ above the ground. Neglecting air resistance,its velocity $(v)$ varies with its height $(y)$ above the ground as:

$A$ boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can,equal to $49\; m s^{-1}$. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of $5\; m s^{-1}$ and the boy again throws the ball up with the maximum speed he can,how long does the ball take to return to his hands?

$A$ balloon rises from rest with a constant acceleration $g/8$. $A$ stone is released from it when it has risen to height $h$. The time taken by the stone to reach the ground is

Two stones begin to fall from rest from the same height,with the second stone starting to fall $\Delta t$ seconds after the first. The distance of separation between the two stones becomes $H$,$t_0$ seconds after the first stone starts its motion. Then $t_0$ is equal to: