Question

The trajectory of a projectile in a vertical plane is $y=\alpha x-\beta x^2$ where $\alpha$ and $\beta$ are constants and x and y are respectively the horizontal and vertical distances of the projectile from the point of projection. The maximum height attended by projectile is

• A. $\frac{\alpha }{\beta }$
• B. $\frac{\alpha }{2\beta }$
• C. $\frac{\beta }{\alpha }$
• D. $\frac{\beta }{2\alpha }$

Answer 1

Answer: (B)

Here, the trajectory of a projectile motion,
$y=\alpha x-\beta x^2\dots (i)$
As, general equation of trajectory of a projectile motion,
$y=tan\theta \cdot x-\frac{g}{2{\left(ucos\theta \right)}^2}\cdot x^2...\left(ii\right)$
Now, comparing Eqs. $(i)$ and $(ii)$, we get
$tan\theta =\alpha$
$\Rightarrow \frac{si{n}^2\theta }{co{s}^2\theta }={\alpha }^2...\left(iii\right)$
and $\beta =\frac{g}{2u^{2}co{s}^2\theta }$
$\Rightarrow 4\beta =\frac{2g}{u^{2}co{s}^2\theta }...\left(iv\right)$
Maximum height of projectile,
$H_{max}=\frac{u^{2}si{n}^2\theta }{2g}$
From Eqs. $(iii)$ and $(iv)$, we get
$\frac{{\alpha }^2}{4\beta }=\frac{u^{2}co{s}^2\theta \times si{n}^2\theta }{2gco{s}^2\theta }=H_{max}$
Hence, the maximum height is $\frac{{\alpha }^2}{4\beta }$.