Question

A ball is projected upwards with a certain speed. Another ball of the same mass is projected at an angle of ${60^0}$ to the vertical with the same initial speed. The ratio of their potential energies at highest points of their journey will be:
(A) 1:1
(B) 2:1
(C) 3:2
(D) 4:1

Answer 1

We know that anybody projected into the air at an angle other than 900with the horizontal is called projectile motion. The maximum height achieved by the ball is directly proportional to the initial velocity and angle made by the projectile body with the surface of earth and inversely proportional to acceleration due to gravity.
Formula used: Maximum height of a projectile is given by,
$H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
Where, u is the initial velocity with which the body is projected (m/s), g is acceleration due to gravity, and $\theta$ is angle made by the projectile with horizontal surface of earth.

Complete step by step answer:
Let us examine the two cases for the ball as shown below,
Given, case (1):
Let,A ball is projected vertically upward with velocity u under gravity g.
Then, the maximum height reached by the ball, ${H_1} = \dfrac{{{u^2}{{\sin }^2}{\theta _1}}}{{2g}}$
Here, ${\theta _1}$ is the angle made by the body with the horizontal that is ${\theta _1} = {90^0}$
We know that, $\sin {90^0} = 1$
Apply the formula of maximum height we get,
${H_1} = \dfrac{{{u^2}{{\sin }^2}{\theta _1}}}{{2g}}$ $= \dfrac{{{u^2}}}{{2g}}$
Therefore, potential energy at the maximum is height ${H_1}$ is given by,
${U_1} = mg{H_1}$
${U_1} = mg\dfrac{{{u^2}}}{{2g}}$ …………………… (1)
Case (2): Let,Another ball of the same mass is projected at an angle of ${60^0}$ to the vertical with the same initial speed then
The maximum height reached by the ball, ${H_2} = \dfrac{{{u^2}{{\sin }^2}{\theta _2}}}{{2g}}$
Here, ${\theta _2}$ is the angle made by the body with the horizontal that is ${\theta _2} = {90^0} - {60^0} = {30^0}$
Then, $\sin {30^0} = \dfrac{1}{2}$
Let us apply formula of maximum height we get,
${H_2} = \dfrac{{{u^2}{{\sin }^2}{\theta _2}}}{{2g}}$ $= \dfrac{{{u^2}}}{{2g}} \times \dfrac{1}{4}$
${H_2} = \dfrac{{{u^2}}}{{8g}}$
Therefore, potential energy at the maximum is height ${H_2}$ is given by,
${U_2} = mg{H_2}$
${U_2} = mg\dfrac{{{u^2}}}{{8g}}$ …………………. (2)
Now divide equation (1) by(2) we get,
$\dfrac{{{U_1}}}{{{U_2}}} = \dfrac{{{u^2}}}{{2g}} \times \dfrac{{8g}}{{{u^2}}}$
$\dfrac{{{U_1}}}{{{U_2}}} = \dfrac{4}{1}$
The ratio of their potential energies at the highest points of their journey will be 4:1.

Therefore, the correct option is (D).

Additional information:
There are three parameters which are related to projectile motion.
(i) Time of flight: Total time to reach the horizontal surface of the projectile is called time of flight. It is the total time for which the projectile remains in air.
(ii) Maximum height: The vertical displacement of the projectile during time of ascent (For a projectile the time to reach maximum height).
(iii) Horizontal range: The horizontal distance covered by the projectile during its motion.
Let a body be projected at O with an initial velocity u that makes an angle $\theta$ with the X-axis.
Due to the fact that two dimensional motions can be treated as two independent rectilinear motions, the projectile motion can be broken up into two separated straight line motions.
Horizontal motion with zero acceleration (that is constant velocity as there is no force in horizontal direction).
Vertical motion with constant downward acceleration = g ( $\because$ it is moving under gravity).

Note:
Maximum height: the vertical displacement of the projectile during time of ascent (for a projectile the time to reach maximum height).
When$\theta = {90^0}$, ${H_{\max }} = \dfrac{{{u^2}}}{{2g}}$this is equal to the maximum height reached by a body projected vertically upwards.