$A$ 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?

$A$ bead of mass $\frac{1}{2} \, kg$ starts from rest from point $A$ to move in a vertical plane along a smooth fixed quarter ring of radius $5 \, m$,under the action of a constant horizontal force $F = 5 \, N$ as shown. The speed of the bead as it reaches point $B$ is ................ $m/s$. [Take $g = 10 \, m/s^2$]

$A$ body falls freely under gravity. Its speed is $v$ when it has lost an amount $U$ of the gravitational potential energy. Then its mass is

$A$ body of mass $m$ is travelling with a velocity $u$. When a constant retarding force $F$ is applied,it comes to rest after travelling a distance $s_{1}$. If the initial velocity is $2u$,with the same force $F$,the distance travelled before it comes to rest is $s_{2}$. Then,

The potential energy of a $1 \ kg$ particle free to move along the $x-$axis is given by: $U(x) = (\frac{x^4}{4} - \frac{x^2}{2}) \ J$. The total mechanical energy of the particle is $2 \ J$. Then,the maximum speed (in $m/s$) is:

$A$ particle is released from a height $H$. At a certain height,its kinetic energy is half of its potential energy with reference to the surface of the earth. The height and speed of the particle at that instant are respectively: