Question

The trajectory of a particle in a vertical plane is $4y=3x-\frac{x^2}{4}$, $x$ and $y$ are the horizontal and vertical distances of the projectile from the point of projection. The angle of projection from the horizontal and the maximum height attained by the particle are

• A. ${45}^{\circ },\frac{3}{4}\text{units}$
• B. ${53}^{\circ },\frac{9}{2}\text{units}$
• C. ${37}^{\circ },\frac{9}{4}\text{units}$
• D. ${60}^{\circ },\frac{9}{16}\text{units}$

Answer 1

The correct option is C ${37}^{\circ },\frac{9}{4}\text{units}$

Given, equation $y=\frac{3}{4}x-\frac{x^2}{16}$
Comparing the above with the equation of projectile
$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$
$\tan \theta =\frac{3}{4}\Rightarrow \theta ={\tan }^{-1}\left(\frac{3}{4}\right)={37}^{\circ }$

(Angle of projection)
We know that maximum height of projectile $\left(H\right)=\frac{u^2{\sin }^2\theta }{2g}$
From equation $\frac{2u^2{\cos }^2\theta }{g}=16$

Multiplying and dividing the equation of maximum height

$\therefore H=\frac{u^2{\sin }^2\theta }{2g}\times \frac{{\tan }^2\theta }{{\tan }^2\theta }=\frac{u^2{\cos }^2\theta }{2g}\times {\tan }^2\theta$
$=\left(\frac{2u^2{\cos }^2\theta }{g}\right)\times \frac{{\tan }^2\theta }{4}$
$=16\times \frac{{\left(\frac{3}{4}\right)}^2}{4}=4\times \frac{9}{16}=\frac{9}{4}\text{units}$

Hence option C is the correct answer