
A ball is rolled off the edge of a horizontal table at a speed of \[4\text{ m/sec}\]. It hits the ground after \[0.4\text{ sec}\]. Which statement given below is true
(A) It hits the ground at a horizontal distance \[1.6\text{ m}\]from the edge of the table
(B) The speed with which it hits the ground is \[0.4\text{ m/sec}\]
(C) Height of the table is \[0.8\text{ m}\]
(D) It hits the ground at an angle of $60{}^\circ $ to the horizontal \[\]
Answer
126.6k+ views
Hint Check the given options according to the data given in the question using Newton’s Equations of Motion. There are three ways to pair these equations- velocity-time, position-time and velocity-position. In this order these are also known as first, second and third equations of motion. The uses of these equations depend on the data given in the question.
FORMULA USED \[s=ut+\dfrac{1}{2}a{{t}^{2}}\],\[v=u+at\]
Where, \[s\]→ displacement in time ‘\[t\]’
\[u\] → initial velocity
\[v\] → final velocity
\[a\] → acceleration
Complete Step by Step solution
Given, \[u=4{}^{m}/{}_{s}\] (initial velocity in horizontal direction)
\[t=0.4s\](time after the ball hits the ground)
So, acceleration will be due to the gravity i.e \[a=g=10m{{s}^{-2}}\]
First checking for option (a): Using kinematic equation of motion,
\[s=ut+\dfrac{1}{2}a{{t}^{2}}\]
For horizontal distance, \[{{s}_{x}}={{u}_{x}}t\] [a=0 because velocity is constant in x-direction]
= 4 x 0.4
= 1.6m
So, our option (a) is correct.
Now, checking for option (b): \[v=u+at\] (Kinematic Equation of Motion)
Initial velocity \[u=0\]in y-direction as the ball is rolled off on a horizontal table.
So, \[v=at\]
= 10 x 0.4
= \[4{}^{m}/{}_{s}\]
So, the speed with which the ball hits the ground is\[4m{{s}^{-1}}\]. So, Option (b) is also correct.
Now, we will check for option (c): For height of the table, we need to find out the vertical displacement.
So, using kinematic equation of motion, \[s=ut+\dfrac{1}{2}a{{t}^{2}}\]
\[s=0+\dfrac{1}{2}a{{t}^{2}}\]
\[u=0\]for y-direction or vertical initial velocity is zero.
So, \[s=\dfrac{1}{2}\times 10\times {{(0.4)}^{2}}\]
= 5 x 0.16
= 0.8 m
So, the height of the table is 0.8m. So, option (c) is also correct.
Now, for option (d): Since, velocity in horizontal direction = \[4m{{s}^{-1}}\]and velocity in vertical direction is also\[4m{{s}^{-1}}\]. So the angle formed by the ball when it hits the ground is \[{{45}^{\circ }}\]to both the vertical and horizontal. So, option (d) is incorrect.
So, finally the correct options are (a), (b) and (c).
Additional Information
While solving numerical problems of bodies moving in a straight horizontal direction, we will consider only the magnitudes of a,v,r and s and take care of their direction by assigning positive or negative signs to the gravity. For ex: +a mean acceleration, -a will mean retardation (or deceleration).
Note Concepts of Kinematics, Equations of motion should be studied properly. One needs to understand the language of question, differentiating between the horizontal and vertical velocities and displacements. Sign conventions must also be taken proper care of.
FORMULA USED \[s=ut+\dfrac{1}{2}a{{t}^{2}}\],\[v=u+at\]
Where, \[s\]→ displacement in time ‘\[t\]’
\[u\] → initial velocity
\[v\] → final velocity
\[a\] → acceleration
Complete Step by Step solution
Given, \[u=4{}^{m}/{}_{s}\] (initial velocity in horizontal direction)
\[t=0.4s\](time after the ball hits the ground)
So, acceleration will be due to the gravity i.e \[a=g=10m{{s}^{-2}}\]
First checking for option (a): Using kinematic equation of motion,
\[s=ut+\dfrac{1}{2}a{{t}^{2}}\]
For horizontal distance, \[{{s}_{x}}={{u}_{x}}t\] [a=0 because velocity is constant in x-direction]
= 4 x 0.4
= 1.6m
So, our option (a) is correct.
Now, checking for option (b): \[v=u+at\] (Kinematic Equation of Motion)
Initial velocity \[u=0\]in y-direction as the ball is rolled off on a horizontal table.
So, \[v=at\]
= 10 x 0.4
= \[4{}^{m}/{}_{s}\]
So, the speed with which the ball hits the ground is\[4m{{s}^{-1}}\]. So, Option (b) is also correct.
Now, we will check for option (c): For height of the table, we need to find out the vertical displacement.
So, using kinematic equation of motion, \[s=ut+\dfrac{1}{2}a{{t}^{2}}\]
\[s=0+\dfrac{1}{2}a{{t}^{2}}\]
\[u=0\]for y-direction or vertical initial velocity is zero.
So, \[s=\dfrac{1}{2}\times 10\times {{(0.4)}^{2}}\]
= 5 x 0.16
= 0.8 m
So, the height of the table is 0.8m. So, option (c) is also correct.
Now, for option (d): Since, velocity in horizontal direction = \[4m{{s}^{-1}}\]and velocity in vertical direction is also\[4m{{s}^{-1}}\]. So the angle formed by the ball when it hits the ground is \[{{45}^{\circ }}\]to both the vertical and horizontal. So, option (d) is incorrect.
So, finally the correct options are (a), (b) and (c).
Additional Information
While solving numerical problems of bodies moving in a straight horizontal direction, we will consider only the magnitudes of a,v,r and s and take care of their direction by assigning positive or negative signs to the gravity. For ex: +a mean acceleration, -a will mean retardation (or deceleration).
Note Concepts of Kinematics, Equations of motion should be studied properly. One needs to understand the language of question, differentiating between the horizontal and vertical velocities and displacements. Sign conventions must also be taken proper care of.
Recently Updated Pages
JEE Main 2023 (April 8th Shift 2) Physics Question Paper with Answer Key

JEE Main 2023 (January 30th Shift 2) Maths Question Paper with Answer Key

JEE Main 2022 (July 25th Shift 2) Physics Question Paper with Answer Key

Classification of Elements and Periodicity in Properties Chapter For JEE Main Chemistry

JEE Main 2023 (January 25th Shift 1) Maths Question Paper with Answer Key

JEE Main 2023 (January 24th Shift 2) Chemistry Question Paper with Answer Key

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 Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!

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

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

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

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