A shooter wants to fire a bullet in such a way ... - JAMB Physics 2006 Question
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?
A
68o
B
53o
C
30o
D
45o
correct option: b
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
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