The horizontal range of a projectile is 4√3 times the maximum height achieved by it. The angle of projection is
Options
(a) 30⁰
(b) 45⁰
(c) 60⁰
(d) 90⁰
Correct Answer:
30⁰
Explanation:
u²sin 2θ/g = 4√3 x (u²sin 2θ/g ) ⇒ tan θ = 1/√3 .·. θ = 30°
The horizontal range of a projectile is 4√3 times the maximum height achieved by it. The angle of projection is
(a) 30⁰
(b) 45⁰
(c) 60⁰
(d) 90⁰
30⁰
u²sin 2θ/g = 4√3 x (u²sin 2θ/g ) ⇒ tan θ = 1/√3 .·. θ = 30°