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(C) The initial velocity is given by $\vec{u} = a \hat{i} + b \hat{j}$,where $u_x = a$ and $u_y = b$.The range $R$ of a projectile is given by $R = \f

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$b = 2a$

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(C) The initial velocity is given by $\vec{u} = a \hat{i} + b \hat{j}$,where $u_x = a$ and $u_y = b$.The range $R$ of a projectile is given by $R = \frac{2 u_x u_y}{g} = \frac{2ab}{g}$.The maximum height $H$ reached by the projectile is given by $H = \frac{u_y^2}{2g} = \frac{b^2}{2g}$.According to the problem,the range is twice the maximum height: $R = 2H$.Substituting the expressions for $R$ and $H$:$\frac{2ab}{g} = 2 \left( \frac{b^2}{2g} ight)$.Simplifying the equation:$\frac{2ab}{g} = \frac{b^2}{g}$.Since $b 
eq 0$,we can divide both sides by $b/g$:$2a = b$ or $b = 2a$.

$A$ projectile is thrown with an initial velocity of $(a \hat{i} + b \hat{j}) \text{ m s}^{-1}$. If the range of the projectile is twice the maximum height reached by it,then

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