Answer
64.8k+ 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
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)