# 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 

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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).