Question

A particle is released from a height $H$. At a certain height, its kinetic energy is two times its potential energy. Height and speed of particle at that instant are
$\begin{align}
  & A)\dfrac{H}{3},\sqrt{\dfrac{2gH}{3}} \\
 & B)\dfrac{H}{3},2\sqrt{\dfrac{gH}{3}} \\
 & C)\dfrac{2H}{3},\sqrt{\dfrac{2gH}{3}} \\
 & D)\dfrac{H}{3},\sqrt{2gH} \\
\end{align}$

Answer 1

Hint: We will use the law of conservation of energy to find the height of the object. The law states that energy can neither be created nor be destroyed but it can be converted from one form to another. After finding the height, we will use that height to find the velocity of the particle at that instant as it is given that kinetic energy at that point is 2 times the potential energy.

Complete step-by-step answer:

Firstly we will find the height at which the kinetic energy is twice of the potential energy. To find this, we will use the law of conservation of energy. The body will be having a total energy when it is placed at height $H$ as,
$T.E=mgH$
This energy will be converted to kinetic and potential energy at the height $X$, where the kinetic energy will be twice its potential energy. i.e.
$\begin{align}
  & T.E=P.E+K.E \\
 & \Rightarrow T.E=P.E+2\left( P.E \right) \\
 & \Rightarrow mgH=mgX+2\left( mgX \right) \\
 & \Rightarrow mgH=3mgX \\
 & \Rightarrow X=\dfrac{H}{3} \\
\end{align}$
Therefore the height at which kinetic energy is twice the potential energy is found to be $\dfrac{H}{3}$.
Now, to find the velocity of the body at this height, we will use the expression of kinetic energy. i.e.
$K.E=\dfrac{1}{2}m{{v}^{2}}$
We know kinetic energy is twice the potential energy,
$\Rightarrow K.E=2mg\dfrac{H}{3}$
Equating these two, we will get the velocity as,
$\begin{align}
  & \dfrac{1}{2}m{{v}^{2}}=\dfrac{2}{3}mgH \\
 & {{v}^{2}}=\dfrac{4}{3}gH \\
 & v=2\sqrt{\dfrac{gH}{3}} \\
\end{align}$
Hence, velocity at height $X$ is found to be $2\sqrt{\dfrac{gH}{3}}$.

So, the correct answer is “Option b”.

Note: We can also find the height by just comparing the ratio of energies. In that case, assume the total energy is divided into 3 equal portions, while the body is reaching ground, it will be having zero potential energy, while it reaches one by third of the initial height, the potential energy will be one by third and kinetic energy will be two by third of the total energy. So, we can conclude the height as $\dfrac{H}{3}$.