JEE Mains · Physics · STD 11 - Motion in plane
- JEE Mains 2022
- Medium
- MCQ
Given below are two statements. One is labelled as Assertion \(A\) and the other is labelled as Reason \(R\). Assertion A :Two identical balls \(A\) and \(B\) thrown with same velocity '\(u\) ' at two different angles with horizontal attained the same range \(R\). If \(A\) and \(B\) reached the maximum height \(h_{1}\) and \(h_{2}\) respectively, then \(R =4 \sqrt{ h _{1} h _{2}}\) Reason R: Product of said heights. \(h _{1} h _{2}=\left(\frac{u^{2} \sin ^{2} \theta}{2 g }\right) \cdot\left(\frac{u^{2} \cos ^{2} \theta}{2 g }\right)\) Choose the \(CORRECT\) answer
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) Both \(A\) and \(R\) are true and \(R\) is the correct explanation of \(A\).
Step-by-step Solution
Detailed explanation
For same range \(\theta_{1}+\theta_{2}=90^{\circ}\)
\(h _{1} h _{2}=\frac{ u ^{2} \sin ^{2} \theta_{1}}{2 g } \times \frac{ u ^{2} \sin ^{2} \theta_{2}}{2 g }\)
\(\theta_{2}=90-\theta_{1}\)
\(h _{1} h _{2}=\frac{ u ^{2} \sin ^{2} \theta_{1}}{2 g } \cdot \frac{ u ^{2} \cos ^{2} \theta_{1}}{2 g }\)
\(=\left[\frac{ u ^{2} \sin \theta_{1} \cos \theta_{1}}{2 g }\right]^{2}\)
\(=\left[\frac{ u ^{2} \sin \theta_{1} \cos \theta_{1}}{2 g } \times \frac{2}{2}\right]^{2}=\frac{ R ^{2}}{16}\)
\(R =4 \sqrt{ h _{1} h _{2}}\)
So \(R\) is correct explanation of \(A\)
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