
A person can throw a stone to a maximum distance of 100 m. The greatest height to which he can throw the stone is:-
(A) 100m
(B) 75m
(C) 50m
(D) 25m
Answer
244.2k+ views
Hint: The maximum distance given, 100m, is the range of projectile motion. The maximum height is achieved when thrown straight up at ${45^ \circ }$ angle. Thus launching angle is $\theta = {45^ \circ }$.
Formula Used: The formulae used in the solution are given here.
Range is given as $R = \dfrac{{{u^2}\sin 2\theta }}{g}$ where $u$ is the initial velocity, $\theta $ is the launch angle and $g$ is the acceleration due to gravity and the maximum height is given by $H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
Complete Step by Step Solution: This is a projectile motion problem. Given that, the range is 100m, we have to find the maximum height that a person can throw a stone.
Range is given as $R = \dfrac{{{u^2}\sin 2\theta }}{g}$ where $u$ is the initial velocity, $\theta $ is the launch angle and $g$ is the acceleration due to gravity.
Since, $R = 100m$, thus $\dfrac{{{u^2}\sin 2\theta }}{g} = 100$ and the acceleration due to gravity $g = 9.8$.
Thus, it can be written that, ${u^2}\sin 2\theta = 100g$ where $g = 9.8$
It is known to us that, the greatest range is achieved using a ${45^ \circ }$ launch angle:
${u^2}\sin \left( {2 \times {{45}^ \circ }} \right) = 980$
$ \Rightarrow {u^2}\sin {90^ \circ } = 980$
We know that, $\sin {90^ \circ } = 1$,
$\therefore {u^2} = 980$
The height is achieved when thrown straight up at ${45^ \circ }$ angle.
Thus, height $H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$.
Since, ${u^2} = 980$, $\theta = {45^ \circ }$ and $g = 9.8$
$H = \dfrac{{980}}{{39.2}}$
$ \Rightarrow H = 25m$
The greatest height to which he can throw the stone is 25m. Hence, the correct answer is Option D.
Note: Some assumptions have to be made in answering the question.
First it is assumed that we can ignore air resistance.
Next, it is assumed that the given range is for the rock being launched and landing at the same height (i.e. on level ground).
Third, it is assumed that when he launches the rock to get maximum height (i.e. when he throws it vertically upward) that he can throw it with the same speed that he throws it to get maximum range (i.e. when he throws it at 45 degrees).
Formula Used: The formulae used in the solution are given here.
Range is given as $R = \dfrac{{{u^2}\sin 2\theta }}{g}$ where $u$ is the initial velocity, $\theta $ is the launch angle and $g$ is the acceleration due to gravity and the maximum height is given by $H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
Complete Step by Step Solution: This is a projectile motion problem. Given that, the range is 100m, we have to find the maximum height that a person can throw a stone.
Range is given as $R = \dfrac{{{u^2}\sin 2\theta }}{g}$ where $u$ is the initial velocity, $\theta $ is the launch angle and $g$ is the acceleration due to gravity.
Since, $R = 100m$, thus $\dfrac{{{u^2}\sin 2\theta }}{g} = 100$ and the acceleration due to gravity $g = 9.8$.
Thus, it can be written that, ${u^2}\sin 2\theta = 100g$ where $g = 9.8$
It is known to us that, the greatest range is achieved using a ${45^ \circ }$ launch angle:
${u^2}\sin \left( {2 \times {{45}^ \circ }} \right) = 980$
$ \Rightarrow {u^2}\sin {90^ \circ } = 980$
We know that, $\sin {90^ \circ } = 1$,
$\therefore {u^2} = 980$
The height is achieved when thrown straight up at ${45^ \circ }$ angle.
Thus, height $H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$.
Since, ${u^2} = 980$, $\theta = {45^ \circ }$ and $g = 9.8$
$H = \dfrac{{980}}{{39.2}}$
$ \Rightarrow H = 25m$
The greatest height to which he can throw the stone is 25m. Hence, the correct answer is Option D.
Note: Some assumptions have to be made in answering the question.
First it is assumed that we can ignore air resistance.
Next, it is assumed that the given range is for the rock being launched and landing at the same height (i.e. on level ground).
Third, it is assumed that when he launches the rock to get maximum height (i.e. when he throws it vertically upward) that he can throw it with the same speed that he throws it to get maximum range (i.e. when he throws it at 45 degrees).
Recently Updated Pages
JEE Main 2026 Session 2 City Intimation Slip & Exam Date: Expected Date, Download Link

JEE Main 2026 Session 2 Application Form: Reopened Registration, Dates & Fees

JEE Main 2026 Session 2 Registration (Reopened): Last Date, Fees, Link & Process

WBJEE 2026 Registration Started: Important Dates Eligibility Syllabus Exam Pattern

Dimensions of Charge: Dimensional Formula, Derivation, SI Units & Examples

How to Calculate Moment of Inertia: Step-by-Step Guide & Formulas

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Understanding the Angle of Deviation in a Prism

Understanding Differential Equations: A Complete Guide

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

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

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

CBSE Notes Class 11 Physics Chapter 4 - Laws of Motion - 2025-26

CBSE Notes Class 11 Physics Chapter 14 - Waves - 2025-26

