$A$ stone is dropped from the top of a building. When it crosses a point $5 \, m$ below the top,another stone starts to fall from a point $25 \, m$ below the top. Both stones reach the bottom of the building simultaneously. The height of the building is ..... $m$.

$A$ person throws balls into the air vertically upward at regular intervals of time of $1 \,s$. The next ball is thrown when the velocity of the ball thrown earlier becomes zero. The height to which the balls rise is (Assume $g = 10 \,m/s^2$) (in $\,m$)

$A$ 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

Two balls are released from the same position at a height of $500 \ m$ above the ground, one after the other, with an interval of $1 \ s$. What is the distance between the two balls when the first ball hits the ground (in $m$)? (Acceleration due to gravity $g = 10 \ m/s^2$)

Two bodies are held separated by $9.8\,m$ vertically one above the other. They are released simultaneously to fall freely under gravity. After $2\,s$,the relative distance between them is $............\,m$

$A$ ball is thrown vertically upwards with a velocity of $20 \; m s^{-1}$ from the top of a multistorey building. The height of the point from where the ball is thrown is $25.0 \; m$ from the ground. How high will the ball rise above the point of projection (in $; m$)? Take $g = 10 \; m s^{-2}$.