$A$ 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 (initially at rest) is released from the top of a tower. The ratio of work done by the force of gravity, in the first, second, and third seconds is

The initial and final velocities of a body projected vertically from the ground are $20 \,ms^{-1}$ and $18 \,ms^{-1}$ respectively. The maximum height reached by the body is (Acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,m$)

$A$ ball is thrown vertically upwards with a velocity of $19.6 \, m/s$ from the top of a tower. The ball strikes the ground after $6 \, s$. The height from the ground up to which the ball can rise is given by $\left(\frac{k}{5}\right) \, m$. The value of $k$ is ..... (use $g = 9.8 \, m/s^2$)

$A$ juggler tosses a ball up in the air with initial speed $u$. At the instant it reaches its maximum height $H$,he tosses up a second ball with the same initial speed. The two balls will collide at a height:

If the water falls from a dam into a turbine wheel $19.6 \ m$ below,then the velocity of water at the turbines is .................. $m/s$ (take $g = 9.8 \ m/s^2$).