Question

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 1

Answer: (D)

$R = H$
$\frac{2 v _{ x } \times v _{ y }}{ g }=\frac{ v _{ y }^2}{2 g }$
$v _{ x }=\frac{ v _{ y }}{4} ; u \cos \theta=\frac{ u \sin \theta}{4}$
$\tan \theta=4$