Question

A body $A$ is projected upwards with velocity ${v_1}$ . Another body $B$ of the same mass is projected at an angle of ${45^ \circ }$ with the horizontal. Both reach the same height. What is the ratio of their initial kinetic energies?
(A) $\dfrac{1}{4}$
(B) $\dfrac{1}{3}$
(C) $\dfrac{1}{2}$
(D) $1$

Answer 1

Hint Energy is always conserved, hence kinetic energy is equal to the potential energy. Find the kinetic energy for the body $A$ and the kinetic energy for the body $B$. Divide both the kinetic energies to find the ratio of the kinetic energy of both the bodies.
Useful formula:
(1) The formula of the kinetic energy is given by
$K = \dfrac{1}{2}m{v^2}$
Where $K$ is the kinetic energy of the body, $m$ is the mass of the body and $v$ is the velocity of the body.
(2) The potential energy is given by
$P = mgh$
Where $P$ is the potential energy of the body, $g$ is the acceleration due to gravity and $h$ is the height of the body.

Complete step by step solution
It is given that the body $B$ is thrown at an angle, $\theta = {45^ \circ }$
It is known that energy is always conserved. Hence the potential energy must be equal to that of the kinetic energy.
${K_A} = {P_A}$ ……………………………….(1)
Substituting the formula (1) and (2) in the above relation.
$\dfrac{1}{2}m{v_A}^2 = mgh$ ……………………………………(2)
For the body $B$ , the movement is inclined. So it must be divided into the horizontal and the vertical component to calculate its kinetic energy. The horizontal component of the velocity of the body $B$ is ${v_B}\cos \theta$ and the vertical component as ${v_B}\sin \theta$ .
Substituting the parameters in the equation (1)
${K_B} = {P_B}$
$\dfrac{1}{2}m{\left( {{v_B}\sin \theta } \right)^2} = mgh$
By substituting the angle of inclination and simplifying the above equation, we get
$\dfrac{1}{2}m\dfrac{{v_B^2}}{2} = mgh$
$\dfrac{1}{2}mv_B^2 = 2mgh$ ………………………………(3)
By dividing the equation (2) and (3) to find the ratio of the kinetic energy.
$\dfrac{{{K_A}}}{{{K_B}}} = \dfrac{{mgh}}{{2mgh}}$
By cancelling the similar terms in the right hand side of the equation, we get
$\dfrac{{{K_A}}}{{{K_B}}} = \dfrac{1}{2}$

Thus the option (C) is correct.

Note: The kinetic energy of the body of the same mass will be greater if it is thrown at some angle to the horizontal rather than vertical. Because lots of energy is required to throw the body upwards against the gravitational force and also the object to move up.