$A$ body is released from the top of a tower of height $h$. It takes $t$ seconds to reach the ground. Where will the body be after time $t/2$ seconds?

$A$ ball is thrown vertically upward. It has a speed of $10 \; m/s$ when it has reached one-half of its maximum height. How high does the ball rise? (Take $g = 10 \; m/s^2$)

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 $g = 10 \, \text{m/s}^2$)

$A$ particle is thrown vertically upwards with a velocity of $4\,ms^{-1}$. The ratio of its accelerations after $1\,s$ and $2\,s$ of its motion is

$A$ body is thrown vertically upwards and reaches its maximum height in $t$ seconds. The total time from the time of projection to reach a point at half of its maximum height while returning (in seconds) is:

$A$ ball is dropped from a bridge that is $45 \ m$ above the water. It falls directly into a boat which is moving with constant velocity. The boat is $12 \ m$ away from the point of impact when the ball is dropped. The speed of the boat is (Take $g=10 \ m \ s^{-2}$) (in $m \ s^{-1}$)