Question

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

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

Answer 1

Answer: (C)

Maximum height, $H=\frac{u^{2} \sin ^{2} \theta}{2 g}$
or $ 80=\frac{u^{2} \sin ^{2} \theta}{2 \times 10}$ or
$u^{2} \sin ^{2} \theta=1600$ or
$u \sin \theta=40 m s ^{-1}$
Horizontal velocity $=u \cos \theta$
As per question, $u \cos \theta=a t=3 \times 30=90 m s ^{-1}$
$\therefore \frac{u \sin \theta}{u \cos \theta}=\frac{40}{90}$ or
$\tan \theta=\frac{4}{9}$ or
$\theta=\tan ^{-1}\left(\frac{4}{9}\right)$