From a height of $h$ above the ground, a ball is projected up at an angle $30^{\circ}$ with the horizontal. If the ball strikes the ground with a speed of $1.25$ times its initial speed of $40 \,ms^{-1}$, the value of $h$ is (acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,m$)

$A$ ball at rest is dropped from a height of $12 \,m$. It loses $25 \%$ of its kinetic energy on striking the ground and bounces back to a height '$h$'. Then the value of '$h$' is (in $\,m$)

$A$ body of mass $2 \,kg$ moving in $X-Y$ plane has a potential energy given by $U = (6x + 8y) \,J$. The body is at rest at the point $(3, 2) \,m$. The work done by the body to reach another position after $2 \,s$ is: (in $\,J$)

$A$ car and a truck are moving with the same kinetic energy. If the same retarding force is applied to both by using brakes,what will be the distance traveled by both vehicles before coming to rest?

What is the velocity of the bob when it reaches point $B$ in $m/s$?

The kinetic energies of two similar cars $A$ and $B$ are $100 \ J$ and $225 \ J$ respectively. On applying brakes,car $A$ stops after $1000 \ m$ and car $B$ stops after $1500 \ m$. If $F_{A}$ and $F_{B}$ are the forces applied by the brakes on cars $A$ and $B$,respectively,then the ratio $F_{A} / F_{B}$ is: