Question

Two second after projection, a projectile is moving at ${30}^o$ above the horizontal, after one more second it is moving horizontally. ($g=10m/se{c}^2)$. The angle of projection is :

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${90}^o$

Answer 1

The correct option is C ${60}^o$

(Refer to Image)

In time interval of $1$ sec

$V_y=at$

$=g\times 1$

$=10m/s$ at $Q$

$V_x=\frac{V_y}{\tan {30}^o}$

$=\frac{10}{1}\sqrt{3}$

$=10\sqrt{3}m/s$

Between $P$ and $R$

$V_y=at$

$=10\times 3$

$=30m/s$

$V_x=10\sqrt{3}m/s$

Angle of projection

$\tan \theta =\frac{V_y}{V_x}$

$=\frac{30}{10\sqrt{3}}$

$=\sqrt{3}$

$\Rightarrow \theta ={\tan }^{-1}\left(\sqrt{3}\right)$

$={60}^o$ .