Question

The maximum height of an oblique projectile is $8\text{m}$. The horizontal range is $24\text{m}$. The horizontal component of the velocity of projection is

• A. $\sqrt{g}$
• B. $2\sqrt{g}$
• C. $3\sqrt{g}$
• D. $4\sqrt{g}$

Answer 1

The correct option is C $3\sqrt{g}$

Let $u$ be the initial velocity of projectile motion. We know that
We know that,
$H=\frac{u^2{\sin }^2\theta }{2g}$
According to question
$\frac{u^2{\sin }^2\theta }{2g}=8$ or $u^2{\sin }^2\theta =16g$
or $u\sin \theta =4\sqrt{g}$
We know that, $R=\frac{u^2\sin 2\theta }{g}$
$24=\frac{2\left(u\sin \theta \right)\left(u\cos \theta \right)}{g}$
$\Rightarrow 12=\frac{\left(4\sqrt{g}\right)\left(u\cos \theta \right)}{g}$
$\Rightarrow u\cos \theta =3\sqrt{g}$
$\therefore$ Horizontal component of velocity $=3\sqrt{g}$