Question

A body is projected with a velocity of 60 ${ms}^{-1}$ at 30${}^o$ to horizontal.

	Column I		Column II
i.	Initial velocity vector	a.	$60\sqrt{3}\hat{i}+40\hat{j}$
ii.	Velocity after 3 s	b.	$30\sqrt{3}\hat{i}+10\hat{j}$
iii.	Displacement after 2 s	c.	$30\sqrt{3}\hat{i}+30\hat{j}$
iv.	Velocity after 2 s	d.	$30\sqrt{3}\hat{i}$

Answer 1

$\text{Step 1: Initial velocity vector[Refer Fig.]}$

$\overrightarrow{u}=u_x\hat{i}+u_y\hat{j}$

$=\left(u\cos {30}^o\right)\hat{i}+\left(u\sin {30}^o\right)\hat{j}$

$=\left(60\cos {30}^o\right)\hat{i}+\left(60\sin {30}^o\right)\hat{j}$

$\overrightarrow{u}=\left(30\sqrt{3}\right)\hat{i}+\left(30\right)\hat{j}m/s\to$Option (c)

$\text{Step 2: Velocity after t= 3 s}$

In $x$ direction $u_x=30\sqrt{3}m/s$

Velocity in x-direction will not change as there is no acceleration in $x$ direction, therefore $v_x=u_x$

In $y$ direction $a_y=-10m/s^2;t=3s$; $u_y=30m/s$

Since acceleration is constant, therefore applying equation of motion

$v_y=u_y+at$

$\Rightarrow$ $v_y=30-10\times 3m/s=0m/s$

$V=30\sqrt{3}\hat{i}m/s\to$Option (d)

$\text{Step 3: Displacement after t= 2 s}$

In x direction: $a_x=0$

$s_x=u_{x}t+\frac{1}{2}{a}_x{t}^2=30\sqrt{3}\times 2\hat{i}+0$ $=60\sqrt{3}\hat{i}m$

In $y$ direction applying equation of motion

$s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$ $=30\times 2-\frac{1}{2}\left(10\right)(2{)}^2$

$s_y=40\hat{j}m$

$\therefore$ $s=(60\sqrt{3}\hat{i}+40\hat{j})m\to$Option (a)

$\text{Step 4: Velocity after 2s}$

In $y$ direction applying equation of motion as the acceleration is constant.

$v_y=u_y+a_{y}t$ $=30-10\times 2m/s$

$=10m/s$

$\therefore$ $v=(30\sqrt{3}\hat{i}+10\hat{j})m/s\to$Option (b)