If a particle takes $t$ seconds less and acquires a velocity of $v \ ms^{-1}$ more in falling through the same distance (starting from rest) on two planets where the accelerations due to gravity are $2g$ and $8g$ respectively,then $v=$

$A$ ball is projected vertically upwards from the ground. It reaches a height $h$ in time $t_1$,continues its motion,and then takes a time $t_2$ to reach the ground. The height $h$ in terms of $g, t_1$,and $t_2$ is ($g =$ acceleration due to gravity).

Two spherical balls having equal masses with a radius of $5 \, \text{cm}$ each are thrown upwards along the same vertical direction at an interval of $3 \, \text{s}$ with the same initial velocity of $35 \, \text{m/s}$. These balls collide at a height of $\ldots \ldots \ldots \, \text{m}$. (Take $g = 10 \, \text{m/s}^2$)

Two balls $A$ and $B$ are thrown with the same velocity $u$ from the top of a tower of height $h$. Ball $A$ is thrown vertically upwards and ball $B$ is thrown vertically downwards. If $t_A$ and $t_B$ are the respective times taken by the balls $A$ and $B$ to reach the ground,then identify the correct statement.

$A$ body is released from the top of a tower $H$ metre high. It takes $t$ seconds to reach the ground. The height of the body $\frac{t}{2}$ seconds after release is

$A$ ball of mass $0.2 \, kg$ is thrown vertically upwards with a velocity of $2 \, m/s$. At the highest point of its trajectory:
$(i)$ What will be the magnitude of its velocity?
$(ii)$ What will be the magnitude of its acceleration?
$(iii)$ What will be the magnitude of the force acting on it? (Take $g = 10 \, m/s^2$.)