JAMB - Physics (2006 - No. 44)
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
Explication
Horizontal range = 3 x maxi. height
=> (U2Sin2θ) / g = (3 x U2sin2θ) / 2g
∴ 4sinθcosθ = 3sin2θ
=> 4cosθ = 3sinθ
∴ sinθ/cosθ = 4/3; => tanθ = 4/3 = 1.3333
θ = tan-1 (1.3333)
= 53o
=> (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