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 the horizontal and vertical distances of the projectile from the point of projection. The maximum height attained is

• A. $\frac{a^2}{4b}$
• B. $\frac{a^2}{8b}$
• C. $\frac{a^2}{2b}$
• D. $\frac{a^2}{2gb}$

Answer 1

The correct option is A $\frac{a^2}{4b}$

The equation of trajectory is, $y=ax-b{x}^2$
For maximum height of projectile, $y$ should be maximum.
$\Rightarrow \frac{dy}{dx}=0$ and $\frac{d^{2}y}{d{x}^2}<0$
$\Rightarrow \frac{dy}{dx}=a-2bx=0$
$\Rightarrow x=\frac{a}{2b}$
$\frac{d^{2}y}{d{x}^2}=-2b<0$
$\therefore$ For $x=\frac{a}{2b}$, $y$ is maximum
Substituting the value of $x=\frac{a}{2b}$ in trajectory equation,
$\Rightarrow y_{\text{max}}=a\left(\frac{a}{2b}\right)-b{\left(\frac{a}{2b}\right)}^2$
$\therefore y_{\text{max}}=\frac{a^2}{4b}$

Alternate solution:
Given equation of trajectory as $y=ax-b{x}^2\Rightarrow y=xa[1-\frac{x}{\left(\frac{a}{b}\right)}]$
Comparing the above equation with equation of trajectory we have
$y=x\tan \theta (1-\frac{x}{R})$
$\Rightarrow \tan \theta =a,R=\frac{a}{b}$
Now we have, $R=\frac{u^2\sin 2\theta }{g}=\frac{2u^2\sin \theta \cos \theta }{g}$
$\Rightarrow R=(\frac{2u^2\sin \theta \cos \theta }{g}\times \frac{\tan \theta }{4})\times \frac{4}{\tan \theta }$
$\Rightarrow R=\left(\frac{u^2{\sin }^2\theta }{2g}\right)\times \frac{4}{\tan \theta }$
$\Rightarrow R=H_{max}\times \frac{4}{\tan \theta }\Rightarrow H_{max}=\frac{R\tan \theta }{4}$
Putting the values of $R$ and $\tan \theta$ into the above relation we get
$H_{max}=\frac{a^2}{4b}$