A projectile is launched at 30° to the horizontal with initial velocity 20 m/s. What is its horizontal range? (g = 10 m/s²)
R=u2sin2θgR = \frac{u^2 \sin 2\theta}{g}R=gu2sin2θ
20 m
40 m
20√3 m
10√3 m
Correct Answer: C — 20√3 m
Explanation:
R=u2sin2θg=202×sin60°10=400×3210=203R = \frac{u^2 \sin 2\theta}{g} = \frac{20^2 \times \sin 60°}{10} = \frac{400 \times \frac{\sqrt{3}}{2}}{10} = 20\sqrt{3}R=gu2sin2θ=10202×sin60°=10400×23=203 m.
