MHT CET · Physics · Motion In Two Dimensions

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

A \(\mathrm{b}=2 \mathrm{a}\)
B \(\mathrm{b}=4 \mathrm{a}\)
C \(\mathrm{b}=\frac{\mathrm{a}}{2}\)
D \(\mathrm{b}=\mathrm{a}\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(\mathrm{b}=2 \mathrm{a}\)

Step-by-step Solution

Detailed explanation

\(\mathrm{u}_{\mathrm{x}}=\mathrm{a}=\) 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, \(\mathrm{R}=\frac{2 \mathrm{u}_{\mathrm{y}} \mathrm{u}_{\mathrm{x}}}{\mathrm{g}}=\frac{2 \mathrm{ba}}{\mathrm{g}}\)
\(
\begin{aligned}
& \because \mathrm{R}=2 \mathrm{H} \\
& \therefore \frac{2 \mathrm{ba}}{\mathrm{g}}=\frac{2 \mathrm{~b}^2}{2 \mathrm{~g}} \\
& \therefore \mathrm{b}=2 \mathrm{a}
\end{aligned}
\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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