Question

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

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