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A projectile is projected with velocity of $25 \mathrm{~m} / \mathrm{s}$ at an angle of with the horizontal. After $\mathrm{t}$ seconds its inclination with horizontal becomes zero. If $\mathrm{R}$ represents horizontal range of the projectile, the valae of $\mathrm{\theta }$ will be :
$\mathrm{\left [ use\: g= 10\, m/s^{2} \right ]}$
Option: 1
$\mathrm{\frac{1}{2} \sin ^{-1} \left [ \frac{5t^{2}}{4R} \right ]}$

Option: 2
$\mathrm{\frac{1}{2} \sin ^{-1} \left [ \frac{4R}{5t^{2}} \right ]}$

Option: 3
$\mathrm{\tan ^{-1}\left[\frac{4 t^{2}}{5R}\right]}$

Option: 4
$\mathrm{\cot ^{-1}\left[\frac{R}{20\, t^{2}}\right]}$

Answers (1)

best_answer

The time at which the inclination of the projectile becomes zero with the horizontal is time of ascent (half of time of flight)

$\mathrm{t= \frac{T}{2}= \frac{u\sin \theta }{g}}$
$\mathrm{R= \frac{u^{2}\times 2\sin \theta \cos \theta }{g}}$
$\mathrm{\frac{R}{t^{2}}= \frac{u^{2}\times 2\sin \theta \cos \theta }{\frac{u^{2}\sin^{2}\theta }{g}}}$
$\mathrm{\frac{R}{t^{2}}= 20\times \cot \theta }$
$\mathrm{\cot \theta = \frac{R}{20\, t^{2}}}$

$\mathrm{ \theta =\cot^{-1}\left ( \frac{R}{20\, t^{2}} \right )}$

The correct option is (4)

Posted by

Divya Prakash Singh

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