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A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3\, ms^{-2}$ for $ 0.5\, minutes$. If the maximum height reached by it is $80\, m$, then the angle of projection is (Take $g = 10\, ms^{-2}$)
${\tan ^{ - 1}}\,\left( 3 \right)$
${\tan ^{ - 1}}\,\left( {\frac{3}{2}} \right)$
${\tan ^{ - 1}}\,\left( {\frac{4}{9}} \right)$
${\sin ^{ - 1}}\,\left( {\frac{4}{9}} \right)$
Solution
$H=\frac{u^{2} \sin ^{2} \theta}{2 g}$
or $\quad 80=\frac{u^{2} \sin ^{2} \theta}{2 \times 10}$
or $\quad \mathrm{u}^{2} \sin ^{2} \theta=1600$
$\mathbf{o r}$ $u \sin \theta=40 \mathrm{ms}^{-1}$
Horizontal velocity $=u \cos \theta$
$=a t$
$=3 \times 30$
$=90 \mathrm{ms}^{-1}$
$\frac{u \sin \theta}{u \cos \theta}=\frac{40}{90}$
or $\tan \theta=\frac{4}{9}$ or $\theta=\tan ^{-1}\left(\frac{4}{9}\right)$
