A projectile is thrown with an initial velocity of v= aî+bĵ, if the range of projectile is double the maximum height reached by it then

The correct option is b = 2a.

Step-by-step Explanation:

Let the initial velocity of the projectile be represented as:
$v=a\hat{i}+b\hat{j}$
Here, the horizontal component of velocity is $u_x=a$ and the vertical component of velocity is $u_y=b$.

The formula for the maximum height ($H$) reached by a projectile is given by:
$H=\frac{u_y^2}{2g}$
Substituting $u_y=b$ into the equation:
$H=\frac{b^2}{2g}$

The formula for the horizontal range ($R$) of the projectile is given by:
$R=\frac{2u_x{u}_y}{g}$
Substituting $u_x=a$ and $u_y=b$ into the equation:
$R=\frac{2ab}{g}$

According to the given condition, the range of the projectile is double its maximum height:
$R=2H$

Substitute the expressions for $R$ and $H$ into the condition:
$\frac{2ab}{g}=2\left(\frac{b^2}{2g}\right)$

Simplify the equation by canceling $2$ on the right-hand side:
$\frac{2ab}{g}=\frac{b^2}{g}$

Now, cancel the acceleration due to gravity ($g$) from the denominators of both sides:
$2ab=b^2$

Since the vertical component of velocity $b\ne 0$ for a projectile, we can divide both sides by $b$:
$2a=b$
Or,
$b=2a$