Question

A body is projected with a velocity $v$ at an angle of projection $\theta$ with the horizontal. The direction of velocity of the body makes angle ${30}^{\circ }$ with the horizontal at $t=2\text{s}$ and then after further $1\text{s}$, it reaches the maximum height. Then
$(g=10{\text{m/s}}^2)$

• A. $v=20\sqrt{3}\text{m/s}$
• B. $\theta ={60}^{\circ }$
• C. $\theta ={30}^{\circ }$
• D. $v=10\sqrt{3}\text{m/s}$

Answer 1

Answer: (A), (B)

$\theta ={60}^{\circ }$

Time of ascent $=2+1=3\text{sec}$
$\Rightarrow \frac{v\sin \theta }{g}=3$
$\Rightarrow v\sin \theta =30$
Now,
$\tan \beta =\frac{v_y}{v_x}$
$\Rightarrow \tan {30}^{\circ }=\frac{v\sin \theta -gt}{v\cos \theta }$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{30-20}{v\cos \theta }\Rightarrow v\cos \theta =10\sqrt{3}$
$\therefore v=\sqrt{v_x^2+v_y^2}=\sqrt{(10\sqrt{3}{)}^2+(30{)}^2}=20\sqrt{3}\text{m/s}$
and $\tan \theta =\frac{v_y}{v_x}=\sqrt{3}\Rightarrow \theta ={60}^{\circ }$