$A$ particle is thrown vertically upwards. It passes the same height at $t_1 = 2 \; s$ and $t_2 = 10 \; s$. The height is:

$A$ ball is thrown upward from the ground with an initial speed of $v$. At the same instant, another ball is dropped from a building of height $20 \,m$. If the balls are at the same height after $0.8 \,s$, then the magnitude of $v$ is (Take, acceleration due to gravity $=10 \,ms^{-2}$) (in $\,ms^{-1}$)

Ball-$1$ is dropped from the top of a building from rest. At the same moment, ball-$2$ is thrown upward towards ball-$1$ with a speed $14 \,m/s$ from a point $21 \,m$ below the top of the building. How far will ball-$1$ have dropped when it passes ball-$2$? (Assume acceleration due to gravity, $g=10 \,m/s^2$.)

$A$ ball is thrown vertically upward from the ground at time, $t=0 \,s$. It passes the top of a tower at $t=3 \,s$ and $2 \,s$ later it reaches its maximum height. The height of the tower is (Acceleration due to gravity, $g=10 \,m/s^2$) (in $\,m$)

$A$ body dropped from the top of a tower covers a distance $7x$ in the last second of its journey,where $x$ is the distance covered in the first second. How much time does it take to reach the ground?

$A$ body is thrown vertically upward at $t = 0$. It is at a height $80 \ m$ at instants $t_1$ and $t_2$. Also,it is at a height $60 \ m$ at instants $t'_1$ and $t'_2$. Then: