Question

A body is projected horizontally from the top of a tower of height $180\, m$ with a velocity of $20 \, ms^{-1}$. If acceleration due to gravity is $10 \, ms^{-2}$ then match the following.

List-I		List-II
A	Velocity of the body after 1 second (in $ms^{-1}$)	I	5
B	Horizontal displacement of the body after 1 second (in meters)	II	20
C	Vertical displacement of the body after 1 second (in meters)	III	10
D	Vertical velocity of the body after 1 second (in $ms^{-1}$)	IV	22.4

• (A) - (IV) ,(B) - (II), (C) - (III) ,(D) - (I)
• (A) - (I), (B) - (II) ,(C) - (III) ,(D) - (IV)
• (A) - (IV) ,(B) - (II) ,(C) - (I) ,(D) - (III)
• (A) - (II), (B) - (IV) ,(C) - (I) ,(D) - (III)

Answer 1

The Correct Option is C

Given, initial horizontal component of velocity,
$u_{x}=20\, m / s$
Initial vertical component of velocity,
$u_{y}=0$
Acceleration due to gravity,
$g=a_{y}=10\, m / s ^{2}$
So, horizontal component of velocity after $1\, s$.
$v_{x} =u_{x}+a_{x} t$
$=20+0=20\, m / s$
and vertical component of velocity after $1\, s$,
$v_{y} =u_{y}+a_{y} t$
$=0+10 \times 1$
$=10\, m / s$
Horizontal displacement after $1\, s$,
$s_{x}=u_{x} t+\frac{1}{2} a_{x} t^{2}$
$=20 \times 1+0=20\, m$
Vertical displacement after $1\, s$,
$s_{y} =u_{y} t+\frac{1}{2} a_{y} t^{2}$
$=0+\frac{1}{2} \times 10 \times(1)^{2}=5\, m$
Resultant velocity after $1\, s$,
$v =\sqrt{v_{x}^{2}+v_{y}^{2}}$
$=\sqrt{(20)^{2}+(10)^{2}}=\sqrt{400+100}=\sqrt{500}$
$=22.36=22.4\, m / s$