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?

$A$ particle slides from the top of a smooth hemispherical surface of radius $R$ which is fixed on a horizontal surface. If it separates from the hemisphere at a height $h$ from the horizontal surface,then the speed of the particle is

$A$ particle of mass $100 \ g$ is thrown vertically upwards with a speed of $5 \ m/s$. The work done by the force of gravity during the time the particle goes up is.....$J$

$A$ particle is released from a height $H$. At a certain height $h$,its kinetic energy is two times its potential energy. The height $h$ and the speed $v$ of the particle at that instant are:

The average force necessary to stop a bullet of mass $20\, g$ moving with a speed of $250\, m/s$,as it penetrates into the wood for a distance of $12\, cm$ is

$A$ bullet moving with a speed of $100 \ m/s$ can just penetrate two planks of equal thickness. Then the number of such planks penetrated by the same bullet when the speed is doubled will be