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