$A$ body thrown vertically upwards from the ground reaches a maximum height $h$. The ratio of the kinetic and potential energies of the body at a height $40 \%$ of $h$ from the ground is

$A$ vertical spring with force constant $k$ is fixed on a table. $A$ ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance $d$. The net work done in the process is

$A$ bullet is fired on a target with velocity $V$. Its velocity decreases from $V$ to $V/2$ when it penetrates $30 \text{ cm}$ in a target. Through what thickness will it penetrate further in the target before coming to rest (in $\text{ cm}$)?

An object is thrown vertically upward with an initial velocity of $4 \ m/s$. At what height (in $m$) will its kinetic energy be halved?

$A$ body of mass $2\, kg$ is thrown up vertically with a kinetic energy of $490\, J$. If the acceleration due to gravity is $9.8\, m/s^2$,then the height at which the kinetic energy of the body becomes half its original value is ............ $m$.

$A$ heavy stone is thrown with speed $v$ from a cliff of height $h$. How and in which direction should it be thrown so that the stone hits the ground with maximum speed?