Question

A projectile thrown with initial velocity $(a\hat{i}+b\hat{j})$ and its range twice maximum height attained by it, then

• A. $b=a/2$
• B. $b=a$
• C. $b=2a$
• D. $b=4a$

Answer 1

Answer: (C)

$b=2a$

According to question,

$R=2H$

$\frac{u^{2}sin2\theta }{g}=2\frac{u^{2}si{n}^2\theta }{2g}$

$2sin\theta .cos\theta =si{n}^2\theta$

$tan\theta =2$. . . . . . . . . .(1)

Projectile velocity,

$\overrightarrow{v}=a\hat{i}+b\hat{j}$

$tan\theta =\frac{b}{a}$. . . . . .(2)

From equation (1) and (2), we get

$\frac{b}{a}=2$

$b=2a$

The correct option is C.