A projectile is thrown with an initial velocity $(\hat{a}+b \hat{j}) \mathrm{m} / \mathrm{s}$, where…
A projectile is thrown with an initial velocity $(\hat{a}+b \hat{j}) \mathrm{m} / \mathrm{s}$, where $\hat{i}$ and $\hat{j}$ are unit vectors along horizontal and vertical directions respectively. If the range of the projectile is twice the maximum height reached by it, then
$\mathrm{b}=2 \mathrm{a}$
$\mathrm{b}=4 \mathrm{a}$
$\mathrm{b}=\frac{\mathrm{a}}{2}$
$\mathrm{b}=\mathrm{a}$
Solution
$\mathrm{u}_{\mathrm{x}}=\mathrm{a}=$ Horizontal component of the velocity
$\mathrm{u}_{\mathrm{y}}=\mathrm{b}=$ Vertical component of the velocity
Maximum height, $\mathrm{H}=\frac{\mathrm{u}_{\mathrm{y}}^2}{2 \mathrm{~g}}=\frac{\mathrm{b}^2}{2 \mathrm{~g}}$
Range, $\mathrm{R}=\frac{2 \mathrm{u}_{\mathrm{y}} \mathrm{u}_{\mathrm{x}}}{\mathrm{g}}=\frac{2 \mathrm{ba}}{\mathrm{g}}$
$\begin{aligned}
& \because \mathrm{R}=2 \mathrm{H} \\
& \therefore \frac{2 \mathrm{ba}}{\mathrm{g}}=\frac{2 \mathrm{~b}^2}{2 \mathrm{~g}} \\
& \therefore \mathrm{b}=2 \mathrm{a}
\end{aligned}$
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