Question

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

Answer 1

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

Using maxima and minima for max height

$\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=a-2bx=0$

$\Rightarrow x=\frac{a}{2b}$

$\therefore$ max height $h=a\left(\frac{a}{2b}\right)-b(\frac{a}{2b}{)}^2$

$h=\frac{a^2}{4b}$ Ans

comparing given equation with equation

of trajectory projectile motion

$y=xTan\theta -\frac{g}{2u^{2}co{s}^2\theta }{x}^2$

$Tan\theta =a$

$\theta =Ta{n}^{-1}\left(a\right)$ Ans