A shooter wants to fire a bullet in such a way that its horizontal range is equal to three times its maximum height. At the what angle should he fire the bullet to achieve this?
- 68o
-
53o
- 30o
- 45o
The correct answer is: B
Explanation
Horizontal range = 3 x maxi. height=> (U2Sin2θ) / g = (3 x U2sin2θ) / 2g
| ∴sin2θ = | 3sin2θ |
| 2 |
| => 2sinθcosθ = | 3sin2θ |
| 2 |
∴ 4sinθcosθ = 3sin2θ
=> 4cosθ = 3sinθ
∴ sinθ/cosθ = 4/3; => tanθ = 4/3 = 1.3333
θ = tan-1 (1.3333)
= 53o