Courses for Kids
Free study material
Offline Centres
Store Icon

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 \[\]

Last updated date: 22nd Jun 2024
Total views: 54.6k
Views today: 2.54k
54.6k+ 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,
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, \[s=ut+\dfrac{1}{2}a{{t}^{2}}\]
\[u=0\]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.