$A$ bullet when fired at a target with a velocity of $100\,m/s$ penetrates $1\,m$ into it. If the bullet is fired at a similar target with a thickness of $0.5\,m$,then it will emerge from it with a velocity of:

$A$ 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 ............ $m$.

The figure shows the vertical section of a frictionless surface. $A$ block of mass $2\, kg$ is released from position $A$. Its kinetic energy $(KE)$ as it reaches position $C$ is .............. $J$.

$A$ force acts on a $3\, g$ particle in such a way that the position of the particle as a function of time is given by $x = 3t - 4t^2 + t^3$,where $x$ is in $meters$ and $t$ is in $seconds$. The work done during the first $4\, seconds$ is ................. $mJ$.

The velocity at which the $6 \text{ kg}$ mass (shown in the figure) strikes the ground when it is released from a height of $6 \text{ m}$ above the ground is . . . . . . $\text{m/s}$. Assume the pulley is massless and the string is light and inextensible. (Take $g = 10 \text{ m/s}^2$)

$A$ bullet of mass $10 \; g$ leaves a rifle at an initial velocity of $1000 \; m/s$ and strikes the earth at the same level with a velocity of $500 \; m/s$. The work done in joule $(J)$ in overcoming the resistance of air will be: