Question

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

• A. $\frac{b^{2}}{2a},tan^{-1}(b)$
• B. $\frac{a^{2}}{b},tan^{-1}(2b)$
• C. $\frac{a^{2}}{4b},tan^{-1}(a)$
• D. $\frac{2a^{2}}{b},tan^{-1}(a)$

Answer 1

Answer: (C)

$y=ax -bx^{2}$
For height or y to be maximum,
$\frac{dy}{dx}=0$ or $a-2bx =0$
or $x=\frac{a}{2b}$
$(i)\,\,\,\, y_{\text{max}} =a (\frac{a}{2d})-b (\frac{a}{2b})^{2}=\frac{a^{2}}{4b}$
$(ii)\,\,\,\, (\frac{dy}{dx})_{x=o} =a=\,\,tan\,\,\theta_{0}$
where $\theta_{0}=$ angle of projection
$\theta_{0} =\,\, tan^{-1} (a)$