Question

A projectile is thrown with an initial velocity $(a\hat{i}+b\hat{j})m{s}^{-1}$, where $\hat{i}$ and $\hat{j}$ are the unit vectors along horizontal and vertical directions respectively. If the range of projectile is twice the maximum height reached by it, then

• A. b = $\frac{a}{2}$
• B. $b=a$
• C. $b=2a$
• D. $b=4a$

Answer 1

Answer: (C)

$b=2a$

Given,

$\overrightarrow{v}=(a\hat{i}+b\hat{j})m/s$

$tan\theta =\frac{b}{a}$. . . . .(1)

If the range of projectile is twice the maximum height reached by particle

$R=2H$

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

$2sin\theta .cos\theta =sin\theta .sin\theta$

$2cos\theta =sin\theta$

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

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

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

$b=2a$

The correct option is C.