A body is projected horizontally from the top of a tower of height $180 \mathrm{~m}$ with a velocity of $20…
A body is projected horizontally from the top of a tower of height $180 \mathrm{~m}$ with a velocity of $20 \mathrm{~ms}^{-1}$. If acceleration due to gravity is $10 \mathrm{~ms}^{-2}$, then match the following.
$\begin{array}{llll}A & B & C & D\end{array}$
IV II III I
I II III IV
IV II I III
II IV I III
Solution
Given, initial horizontal component of velocity,
$
u_x=20 \mathrm{~m} / \mathrm{s} \text {. }
$
Initial vertical component of velocity,
$
u_y=0
$
Acceleration due to gravity,
$
g=a_y=10 \mathrm{~m} / \mathrm{s}^2
$
So, horizontal component of velocity after $1 \mathrm{~s}$,
$
\begin{aligned}
v_x & =u_x+a_x t \\
& =20+0=20 \mathrm{~m} / \mathrm{s}
\end{aligned}
$
and vertical component of velocity after $1 \mathrm{~s}$,
$
\begin{aligned}
v_y & =u_y+a_y t \\
& =0+10 \times 1 \\
& =10 \mathrm{~m} / \mathrm{s}
\end{aligned}
$
Horizontal displacement after $1 \mathrm{~s}$,
$
\begin{aligned}
s_X & =u_x t+\frac{1}{2} a_X t^2 \\
& =20 \times 1+0=20 \mathrm{~m}
\end{aligned}
$
Vertical displacement after $1 \mathrm{~s}$,
$
\begin{aligned}
s_y & =u_y t+\frac{1}{2} a_y t^2 \\
& =0+\frac{1}{2} \times 10 \times(1)^2=5 \mathrm{~m}
\end{aligned}
$
Resultant velocity after $1 \mathrm{~s}$,
$
\begin{aligned}
v= & \sqrt{v_x^2+v_y^2} \\
= & \sqrt{(20)^2+(10)^2}=\sqrt{400+100}=\sqrt{500} \\
& =22.36=22.4 \mathrm{~m} / \mathrm{s},
\end{aligned}
$
So, the correct option is (c)
