$A$ body is thrown vertically upwards and reaches a maximum height $h$. What is the ratio of its kinetic energy to its potential energy at a height of $\frac{3h}{4}$?

$A$ particle is released from height $S$ above the surface of the earth. At a certain height,its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively:

$A$ particle of mass $m$ is moving along a circle of radius $R$ such that its tangential acceleration $a_t$ varies with distance covered $x$ as $a_t = \alpha x^2$,where $\alpha$ is a constant. The kinetic energy $K$ of the particle varies with the distance as $K = \beta x^c$,where $\beta$ and $c$ are constants. The values of $\beta$ and $c$ are:

$A$ 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:

$A$ 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:

$A$ simple pendulum is released from point $A$ as shown in the figure. If $m$ and $l$ represent the mass of the bob and the length of the pendulum respectively,the gain in kinetic energy at point $B$ is: