Question

The equation of motion of a projectile is given by $y=12x-\frac{3}{4}{x}^2$. The horizontal component of velocity is $3\text{m/s}$. Given that $g=10{\text{m/s}}^2$, what is the range of the projectile?

• A. $12.4\text{m}$
• B. $21.6\text{m}$
• C. $30.6\text{m}$
• D. $36.0\text{m}$

Answer 1

The correct option is B $21.6\text{m}$

Given,
$y=12x-\frac{3}{4}{x}^2$
$\Rightarrow \frac{dy}{dt}=12\frac{dx}{dt}-\frac{3}{2}x\frac{dx}{dt}$
At $x=0;\frac{dy}{dt}=12\frac{dx}{dt}$

If $\theta$ be the angle of projection, then
$\frac{dy/dt}{dx/dt}=12=\tan \theta$

Also, if $u=$ initial velocity, then $u\cos \theta =3$ [given]

Hence, $\tan \theta \times u\cos \theta =36$
or $u\sin \theta =36$

Range, $R=\frac{u^2\sin 2\theta }{g}=\frac{2u^2\sin \theta \cos \theta }{g}$
$=\frac{2\left(u\sin \theta \right)\left(u\cos \theta \right)}{10}=\frac{2\times 36\times 3}{10}=21.6\text{m}$
Hence, the correct answer is option (b).