$A$ ball is released from the top of a tower of height $H \ m$. It takes $T \ s$ to reach the ground. The height of the ball from the ground after $\frac{T}{4} \ s$ is

$A$ body falls freely for $10 \ s$. What is its average velocity during this journey (in $m/s$)? (Take $g = 10 \ m/s^2$)

$A$ ball is thrown vertically upwards with a speed of $10 \ m/s$ from the top of a tower $200 \ m$ high and another is thrown vertically downwards with the same speed simultaneously. The time difference between them in reaching the ground (in $s$) is: $(g = 10 \ m/s^2)$

Two balls $A$ and $B$ are thrown with the same velocity $u$ from the top of a tower. Ball $A$ is thrown vertically upwards and ball $B$ is thrown vertically downwards. Choose the correct statement.

$A$ body is thrown vertically up from the ground. It reaches a maximum height of $100\,m$ in $5\,sec$. After what time will it reach the ground from the maximum height position?

$A$ body is falling freely from the top of a tower of height $125 \ m$. The distance covered by the body during the last second of its motion is $x \%$ of the height of the tower. Then $x$ is (Acceleration due to gravity $= 10 \ m/s^2$)