JEE Mains · Physics · STD 11 - 3.2 motion in plane
A ball is projected from the ground with a speed \(15 \,ms ^{-1}\) at an angle \(\theta\) with horizontal so that its range and maximum height are equal, then \(tan\,\theta\) will be equal to
- A \(\frac{1}{4}\)
- B \(\frac{1}{2}\)
- C \(2\)
- D \(4\)
Answer & Solution
Correct Answer
(D) \(4\)
Step-by-step Solution
Detailed explanation
\(R = H\) \(\frac{2 v _{ x } \times v _{ y }}{ g }=\frac{ v _{ y }^{2}}{2 g }\) \(v _{ x }=\frac{ v _{ y }}{4} ; u \cos \theta=\frac{ u \sin \theta}{4}\) \(\tan \theta=4\)
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