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The trajectory of a projectile in a vertical plane is $y=ax-bx^2$ where $a$ and $b$ are constants. The maximum height attained by the particle and the angle of projection from horizontal are

\[(a)\;\frac{b^2}{2a}, \tan ^{-1} (b)\quad (b)\;\frac{a^2}{4b},\tan ^{-1}(a) \quad (c)\;\frac{a^2}{b},\tan ^{-1} (2a) \quad (d)\;\frac{2a^2}{b},\tan ^{-1}(a)\]

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2 Answers

for maximum height
Therefore $y_{\large max}=a \bigg(\large\frac{a}{2b}\bigg)$$-b\bigg(\large\frac{a}{2b}\bigg)^2=\frac{a^2}{4b}$
angle of projection $\tan {\theta}=> at \;x=0$
$\quad\qquad\qquad=a-2bx \bigg|_{x=0}$
Therfore $\tan \theta =a$
$\qquad \theta=\tan ^{-1} (a)$
Hence b is the correct answer. 


answered Jul 1, 2013 by meena.p
edited Jan 26, 2014 by meena.p
answer is (B)

a^2/4b, tan^-1
answered Apr 24 by vipul.saini442

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