Question

A particle is projected at an angle with the horizontal such that it follows a trajectory given by the equation $y=6x–2x^2.$ Find the maximum height attained by it.

• A. $2.5m$
• B. $5.2m$
• C. $4.5m$
• D. $6m$

Answer 1

The correct option is C $4.5m$

Trajectory of a projectile is given by the eqn
$y=x\tan \theta (1-\frac{x}{R})$
If $y=ax-b{x}^2,$ then $\tan \theta =a$
and $b=\frac{\tan \theta }{R}\Rightarrow R=\frac{a}{b}$
For maximum height $H,$ we have
maximum height, $H=\frac{u^2{\sin }^2\theta }{2g}$ and Range $R=\frac{u^2\sin 2\theta }{g}$
So, we have the ratio $\frac{H}{R}=\frac{\frac{u^2{\sin }^2\theta }{2g}}{\frac{u^2\sin 2\theta }{g}}$
$\Rightarrow \frac{4H}{R}=\tan \theta \text{Since,}y=6x–2x^2\Rightarrow H=\frac{1}{4}R\tan \theta =\frac{a^2}{4b}=\frac{6^2}{4\times 2}=4.5m$