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JEE Mains · Physics · STD 11 - Motion in plane
  • JEE Mains 2023
  • Medium
  • MCQ

Two objects are projected with same velocity ' \(u\) ' however at different angles \(\alpha\) and \(\beta\) with the horizontal. If \(\alpha+\beta=90^{\circ}\), the ratio of horizontal range of the first object to the \(2^{\text {nd }}\) object will be :

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Answer & Solution

Correct Answer

(D) \(1: 1\)

Step-by-step Solution

Detailed explanation

\(\text { Range }=\frac{ u ^2 \sin 2 \theta}{ g }\)

Range for projection angle " \(\alpha\) "

\(R _1=\frac{ u ^2 \sin 2 \alpha}{ g }\)

Range for projection angle " \(\beta\) "

\(R _2=\frac{ u ^2 \sin 2 \beta}{ g }\)

\(\alpha+\beta=90^{\circ}(\text { Given })\)

\(\Rightarrow \beta=90^{\circ}-\alpha\)

\(R _2=\frac{ u ^2 \sin 2\left(90^{\circ}-\alpha\right)}{ g }\)

\(R _2=\frac{ u ^2 \sin \left(180^{\circ}-2 \alpha\right)}{ g }\)

\(R _2=\frac{ u ^2 \sin 2 \alpha}{ g }\)

\(\Rightarrow \frac{ R _1}{ R _2}=\frac{\left(\frac{ u ^2 \sin 2 \alpha}{ g }\right)}{\left(\frac{ u ^2 \sin 2 \alpha}{ g }\right)}=\frac{1}{1}\)

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