A projectile has initially the same horizontal velocity as it wou… - Vidyadip

MCQ

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 $g = 10\, ms^{-2}$)

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

✓

Answer

Correct option: C.

${\tan ^{ - 1}}\,\left( {\frac{4}{9}} \right)$

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

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

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