Question

A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3{\text{m/s}}^2$ for $0.5$ min. If the maximum height reached by it is $80\text{m}$, then the angle of projection is ($g=10{\text{m/s}}^2$)

• A. ${\tan }^{-1}3$
• B. ${\tan }^{-1}\frac{3}{2}$
• C. ${\tan }^{-1}\frac{4}{9}$
• D. ${\sin }^{-1}\frac{4}{9}$

Answer 1

The correct option is C ${\tan }^{-1}\frac{4}{9}$

We know, $H=\frac{u^2{\sin }^2\theta }{2g}$
$\Rightarrow \frac{u^2{\sin }^2\theta }{2\times 10}=80$
$\Rightarrow u\sin \theta =40{\text{ms}}^{-1}$ ... (1)
Horizontal velocity $=u\cos \theta$ $=u+at=0+3\times 30=90$
$\because t=0.5$ min $=30\text{s}$
$\Rightarrow u\cos \theta =90{\text{ms}}^{-1}$ ... (2)
From (1) and (2),
$\frac{u\sin \theta }{u\cos \theta }=\frac{40}{90}$
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{4}{9}\right)$