Question

A ball is projected from the ground with a speed 15 ms1 at an angle θ with horizontal so that its range and maximum height are equal, then ‘tan θ’ will be equal to

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. 2
• D. 4

Answer 1

Answer: (D)

4

Answer 2

∵ tanθ=$\frac{4H}{R}$

⇒ tanθ = 4 × 1

⇒ tanθ = 4"