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Motion in a Plane

Section: Physics

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Question : 3 of 28

Marks: +1, -0

The angle of projection of a projectile whose path is shown in the given figure is

[AP EAPCET 23 May 2025 S1]

$\tan^{-1}(1)$
$\tan^{-1}\left(\frac{8}{3}\right)$
$\tan^{-1}\left(\frac{4}{3}\right)$
$\tan^{-1}\left(\frac{5}{3}\right)$

Solution:

Range of projectile

R=40\,\text{m}

R\;=40=\;\frac{u^{2}\sin 2\theta}{g}

\Rightarrow u^{2}\;=\;\frac{40g}{\sin 2\theta}

and from equation of trajectory

\;Y=x \tan\theta-\;\frac{1}{2} g \;\frac{x^{2}}{u^{2}\cos^{2}\theta}

\;Y=x \tan\theta-\;\frac{1}{2} \;\frac{g x^{2}\sin 2\theta}{40g\times\cos^{2}\theta}

\;20=30 \tan\theta-\;\frac{1}{80} \times \;\frac{900 \times 2 \sin\theta \cos\theta}{\cos^{2}\theta}

\;20=30 \tan\theta-\;\frac{1800}{80} \tan\theta

\;20=30 \tan\theta-\;\frac{45}{2} \tan\theta

\;20=\;\frac{15 \tan\theta}{2}

\;\Rightarrow 15 \tan\theta=40

\;\;\; \tan\theta=\;\frac{40}{15}=\;\frac{8}{3}

\;\theta=\tan^{-1}\left(\frac{8}{3}\right)

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