Question

The trajectory of a projectile in a vertical plane is $y=ax-b{x}^2$, 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. $\frac{b^2}{2a},ta{n}^{-1}(b)$
• B. $\frac{a^2}{b},ta{n}^{-1}(2b)$
• C. $\frac{a^2}{4b},ta{n}^{-1}(a)$
• D. $\frac{2a^2}{b},ta{n}^{-1}(a)$

Answer 1

Answer: (C)

$y=ax-b{x}^2$

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

${\theta }_0=ta{n}^{-1}(a)$