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Question: Two identical balls are projected, one vertically up and the other one at an angle of \({30^ \circ }...

Two identical balls are projected, one vertically up and the other one at an angle of 30{30^ \circ } to the horizontal with the same initial speed. The potential energy at the respective highest point will be in the ratio:
A) 3:43:4
B) 4:34:3
C) 4:14:1
D) 1:41:4

Explanation

Solution

We know that Potential Energy (P.E.) at a height (vertical) from the ground is equal to mghmgh, where mm = mass of the object, gg = acceleration due to gravity and hh = vertical height from the ground. We also know from the equation of motion that h=u2sin2θ2gh = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}, where uu = initial speed, θ\theta = angle made with the ground.

Complete step by step answer:
For the first ball, P.E. = mgh1mg{h_1}, where h1{h_1} = height from the ground at the highest point for the first ball.
Thus, h1=u12sin2θ12g{h_1} = \dfrac{{{u_1}^2{{\sin }^2}{\theta _1}}}{{2g}}, θ1{\theta _1} = angle with horizontal for first ball = 90{90^ \circ }
Similarly for the second ball, P.E. = mgh2mg{h_2}, where h2{h_2} = height from the ground at the highest point for the second ball.
Thus h2=u22sin2θ22g{h_2} = \dfrac{{{u_2}^2{{\sin }^2}{\theta _2}}}{{2g}}, θ2{\theta _2} = angle with horizontal for second ball = 30{30^ \circ }

Therefore,
Since the initial speed and acceleration due to gravity is same for the two balls,
P.E.1P.E{._1} = potential energy of first ball at highest point
P.E.2P.E{._2} = potential energy of second ball at highest point
Their ratio P.E.1P.E.2\dfrac{{P.E{._1}}}{{P.E{._2}}}
= h1h2\dfrac{{{h_1}}}{{{h_2}}}
= sin2θ1sin2θ2\dfrac{{{{\sin }^2}{\theta _1}}}{{{{\sin }^2}{\theta _2}}}
= sin290sin230\dfrac{{{{\sin }^2}{{90}^ \circ }}}{{{{\sin }^2}{{30}^ \circ }}}
= 41\dfrac{4}{1}
= 4:14:1
Therefore, the answer is option CC.

Note: Be careful about calculation mistakes that might creep in while performing substitutions. Remember the equation of motion and the potential energy formula and how to use them together by eliminating the same variables. We also see that height is directly proportional to sin2θ{\sin ^2}\theta if the initial speed is the same. This direct result can be used in questions to calculate ratios of height by just taking ratio of sine of the angles. Potential Energy is always calculated for vertical height only since it depends only on the vertical height and not on the horizontal displacement as we can see in the question that one ball goes vertically up and other follows projectile motion (both vertical and horizontal), but only vertical displacement of both is taken into account to solve the problem.