$A$ body at rest falls through a height $h$ with velocity $V$. If it has to fall down further for its velocity to become $3V$,the distance travelled in that interval is: (in $h$)

With what velocity should a ball be projected vertically so that the distance covered by it in the $5^{th}$ second is twice the distance it covers in its $6^{th}$ second? (Take $g = 10 \, m/s^2$)

$A$ 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

$A$ ball is dropped from the roof of a tower of height $h$. The total distance covered by it in the last second of its motion is equal to the distance covered by it in the first three seconds. The value of $h$ in meters is $(g = 10\,m/s^2)$.

When a ball is thrown up vertically with velocity $V_o$,it reaches a maximum height of $h$. If one wishes to triple the maximum height,then the ball should be thrown with what velocity?

Due to the presence of air resistance,if a body dropped from a height of $20 \ m$ reaches the ground with a speed of $18 \ m/s$,then the time taken by the body to reach the ground is nearly (in $s$)