
A stone is projected from the ground with a velocity of $14m/s$. One second later it clears a wall $2m$ high. The angle of projection is $\left( {g = 10m{s^{ - 2}}} \right)$
A. ${45^ \circ }$
B. ${30^ \circ }$
C. ${60^ \circ }$
D. ${15^ \circ }$
Answer
123.6k+ views
Hint First divide the velocity into its two components and then substitute the vertical component in the third equation of motion. Solving this equation we can establish the angle of projection of the stone.
Formula used:
$S = ut - \dfrac{1}{2}g{t^2}$ where $S$ is the vertical distance, $u$ is the velocity, $t$ is the time taken and $g$ is the acceleration due to gravity.
Complete step by step answer
We can solve this problem by using Newton's laws of motion. There are two types of systems in classical mechanics- dynamic and kinematic. Dynamics essentially describes the motion of a system considering the forces acting on the body whereas kinematics describes the motion of a system without considering the action of forces acting on the body. In this particular problem, we are dealing with a dynamic system.
When a stone is projected from the ground with a given velocity, we can divide its velocity into two components- a $\sin $ component which is the vertical component and a $\cos $ component which is the horizontal component with respect to the ground. It is given that the stone clears a wall of height $2m$ so we can use newton’s third equation of motion
$
S = ut - \dfrac{1}{2}a{t^2} \\
\Rightarrow 2 = t14\sin \theta - \dfrac{1}{2}g{t^2} \\
\Rightarrow 2 = \left( {14\sin \theta \times 1} \right) - \dfrac{1}{2} \times 10 \times {1^2} \\
\Rightarrow 2 = 14\sin \theta - 5 \\
\Rightarrow 14\sin \theta = 7 \\
\Rightarrow \sin \theta = \dfrac{1}{2} \\
\Rightarrow \theta = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) \\
\Rightarrow \theta = {30^ \circ } \\
$
Therefore, the angle of projection of the stone with respect to the ground is ${30^ \circ }$ So, the correct option is B.
Note Typically the third equation of motion is described as $S = ut + \dfrac{1}{2}a{t^2}$ but in this particular problem we use the negative sign in this equation as the stone is thrown upwards so the force applied is opposite to the acceleration to the gravity acting on the stone.
Formula used:
$S = ut - \dfrac{1}{2}g{t^2}$ where $S$ is the vertical distance, $u$ is the velocity, $t$ is the time taken and $g$ is the acceleration due to gravity.
Complete step by step answer
We can solve this problem by using Newton's laws of motion. There are two types of systems in classical mechanics- dynamic and kinematic. Dynamics essentially describes the motion of a system considering the forces acting on the body whereas kinematics describes the motion of a system without considering the action of forces acting on the body. In this particular problem, we are dealing with a dynamic system.
When a stone is projected from the ground with a given velocity, we can divide its velocity into two components- a $\sin $ component which is the vertical component and a $\cos $ component which is the horizontal component with respect to the ground. It is given that the stone clears a wall of height $2m$ so we can use newton’s third equation of motion
$
S = ut - \dfrac{1}{2}a{t^2} \\
\Rightarrow 2 = t14\sin \theta - \dfrac{1}{2}g{t^2} \\
\Rightarrow 2 = \left( {14\sin \theta \times 1} \right) - \dfrac{1}{2} \times 10 \times {1^2} \\
\Rightarrow 2 = 14\sin \theta - 5 \\
\Rightarrow 14\sin \theta = 7 \\
\Rightarrow \sin \theta = \dfrac{1}{2} \\
\Rightarrow \theta = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) \\
\Rightarrow \theta = {30^ \circ } \\
$
Therefore, the angle of projection of the stone with respect to the ground is ${30^ \circ }$ So, the correct option is B.
Note Typically the third equation of motion is described as $S = ut + \dfrac{1}{2}a{t^2}$ but in this particular problem we use the negative sign in this equation as the stone is thrown upwards so the force applied is opposite to the acceleration to the gravity acting on the stone.
Recently Updated Pages
The ratio of the diameters of two metallic rods of class 11 physics JEE_Main

What is the difference between Conduction and conv class 11 physics JEE_Main

Mark the correct statements about the friction between class 11 physics JEE_Main

Find the acceleration of the wedge towards the right class 11 physics JEE_Main

A standing wave is formed by the superposition of two class 11 physics JEE_Main

Derive an expression for work done by the gas in an class 11 physics JEE_Main

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

JEE Main Login 2045: Step-by-Step Instructions and Details

Class 11 JEE Main Physics Mock Test 2025

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

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

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

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

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