Question

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

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

Answer 1

The correct option is C $ta{n}^{-1}\left(\frac{4}{9}\right)$

$H=\frac{u^2{\sin }^2\theta }{2g}$

or $80=\frac{u^2{\sin }^2\theta }{2\times 10}$

or $u^2{\sin }^2\theta =1600$

or $u\sin \theta =40m{s}^{-1}$

$\frac{u\sin \theta }{u\cos \theta }=\frac{40}{90}$

or $\tan \theta =\frac{4}{9}$

or $\theta ={\tan }^{-1}\left(\frac{4}{9}\right)$.