# 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

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

## 2 Answers

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

answered Jul 1, 2013 by
edited Jan 26, 2014 by meena.p

answer is (B)

a^2/4b, tan^-1
answered Apr 24

1 answer

1 answer

1 answer

1 answer

1 answer

1 answer

1 answer