Question

A body is projected with velocity u at an angle of projection $\theta$ with the horizontal. The direction of velocity of the body makes angle ${30}^{\circ }$ with the horizontal at $t=2s$ and then after $1s$ it reaches the maximum height. Then:

• A. $u=20\sqrt{3}m{s}^{-1}$
• B. $\theta ={60}^{\circ }$
• C. $\theta ={30}^{\circ }$
• D. $u=10\sqrt{3}m{s}^{-1}$

Answer 1

Answer: (A), (B)

The correct options are
A $u=20\sqrt{3}m{s}^{-1}$
B $\theta ={60}^{\circ }$

Time of ascent $=2+1=3s$
$\Rightarrow \frac{u\sin \theta }{g}=3\Rightarrow u\sin \theta =30$
And $\tan \beta =\frac{u\sin \theta -gt}{u\cos \theta }\Rightarrow \tan {30}^{\circ }=\frac{30-10\times 2}{u\cos \theta }$
$\Rightarrow u\cos \theta =10\sqrt{3}$
From here $u=\sqrt{(10\sqrt{3}{)}^2+{10}^2}=20\sqrt{3}m{s}^{-1}$
And $\tan \theta =\frac{30}{10\sqrt{3}}=\sqrt{3}\Rightarrow \theta ={60}^{\circ }$.