Question

The maximum height of an oblique projectile is $8\text{m}$. The horizontal range is $24\text{m}$. The vertical 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 D $4\sqrt{g}$

Let $u$ be the initial velocity of projectile motion.
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 $u\sin \theta =4\sqrt{g}$
$\therefore$ Vertical component of velocity of projection $=4\sqrt{g}$