Question

A projectile is thrown with an initial velocity of $(x\hat{i}+y\hat{j})\text{m/s}$. If the range of the projectile is double the maximum height reached by it, then

• A. $y=2x$
• B. $y=3x$
• C. $y=x$
• D. $y=4x$

Answer 1

The correct option is A $y=2x$

We know that, the maximum height formula is,
$H_{max}=\frac{u^2{\sin }^2\theta }{2g}...\left(ii\right)$, and the range of the projectile is given by,
$R=\frac{u^2\sin 2\theta }{g}...\left(ii\right)$
It is also given that the range of the projectile is double the maximum height reached

$R=2H$
$\frac{u^2\sin 2\theta }{g}=2\times \frac{u^2{\sin }^2\theta }{2g}$, by simplifying we get,

$\Rightarrow \tan \theta =2$
$\Rightarrow \frac{v_y}{v_x}=2=\frac{y}{x}\Rightarrow y=2x$

Hence option A is the correct answer