MCQ
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$2$
- ✓$4$
✓
Answer
Correct option: D.
$4$
d
$R = H$
$R = H$
$\frac{2 v _{ x } \times v _{ y }}{ g }=\frac{ v _{ y }^{2}}{2 g }$
$v _{ x }=\frac{ v _{ y }}{4} ; u \cos \theta=\frac{ u \sin \theta}{4}$
$\tan \theta=4$
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