
Two identical balls are projected, one vertically up and the other one at an angle of ${30^ \circ }$ to the horizontal with the same initial speed. The potential energy at the respective highest point will be in the ratio:
A) $3:4$
B) $4:3$
C) $4:1$
D) $1:4$
Answer
134.4k+ views
Hint: We know that Potential Energy (P.E.) at a height (vertical) from the ground is equal to $mgh$, where $m$ = mass of the object, $g$ = acceleration due to gravity and $h$ = vertical height from the ground. We also know from the equation of motion that $h = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$, where $u$ = initial speed, $\theta $ = angle made with the ground.
Complete step by step answer:
For the first ball, P.E. = \[mg{h_1}\], where \[{h_1}\] = height from the ground at the highest point for the first ball.
Thus, ${h_1} = \dfrac{{{u_1}^2{{\sin }^2}{\theta _1}}}{{2g}}$, ${\theta _1}$ = angle with horizontal for first ball = ${90^ \circ }$
Similarly for the second ball, P.E. = \[mg{h_2}\], where \[{h_2}\] = height from the ground at the highest point for the second ball.
Thus ${h_2} = \dfrac{{{u_2}^2{{\sin }^2}{\theta _2}}}{{2g}}$, ${\theta _2}$ = angle with horizontal for second ball = ${30^ \circ }$
Therefore,
Since the initial speed and acceleration due to gravity is same for the two balls,
$P.E{._1}$ = potential energy of first ball at highest point
$P.E{._2}$ = potential energy of second ball at highest point
Their ratio $\dfrac{{P.E{._1}}}{{P.E{._2}}}$
= $\dfrac{{{h_1}}}{{{h_2}}}$
= $\dfrac{{{{\sin }^2}{\theta _1}}}{{{{\sin }^2}{\theta _2}}}$
= $\dfrac{{{{\sin }^2}{{90}^ \circ }}}{{{{\sin }^2}{{30}^ \circ }}}$
= $\dfrac{4}{1}$
= $4:1$
Therefore, the answer is option $C$.
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 ${\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.
Complete step by step answer:
For the first ball, P.E. = \[mg{h_1}\], where \[{h_1}\] = height from the ground at the highest point for the first ball.
Thus, ${h_1} = \dfrac{{{u_1}^2{{\sin }^2}{\theta _1}}}{{2g}}$, ${\theta _1}$ = angle with horizontal for first ball = ${90^ \circ }$
Similarly for the second ball, P.E. = \[mg{h_2}\], where \[{h_2}\] = height from the ground at the highest point for the second ball.
Thus ${h_2} = \dfrac{{{u_2}^2{{\sin }^2}{\theta _2}}}{{2g}}$, ${\theta _2}$ = angle with horizontal for second ball = ${30^ \circ }$
Therefore,
Since the initial speed and acceleration due to gravity is same for the two balls,
$P.E{._1}$ = potential energy of first ball at highest point
$P.E{._2}$ = potential energy of second ball at highest point
Their ratio $\dfrac{{P.E{._1}}}{{P.E{._2}}}$
= $\dfrac{{{h_1}}}{{{h_2}}}$
= $\dfrac{{{{\sin }^2}{\theta _1}}}{{{{\sin }^2}{\theta _2}}}$
= $\dfrac{{{{\sin }^2}{{90}^ \circ }}}{{{{\sin }^2}{{30}^ \circ }}}$
= $\dfrac{4}{1}$
= $4:1$
Therefore, the answer is option $C$.
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 ${\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.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

What are examples of Chemical Properties class 10 chemistry JEE_Main

JEE Main 2025 Session 2 Schedule Released – Check Important Details Here!

JEE Main 2025 Session 2 Admit Card – Release Date & Direct Download Link

JEE Main 2025 Session 2 Registration (Closed) - Link, Last Date & Fees

JEE Mains Result 2025 NTA NIC – Check Your Score Now!

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Degree of Dissociation and Its Formula With Solved Example for JEE

A body is falling from a height h After it has fallen class 11 physics JEE_Main

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion In A Plane: Line Class 11 Notes: CBSE Physics Chapter 3
