A projectile is launched from ground level with an initial velocity of 30 m/s at an angle of 60° above the horizontal. What is the horizontal range of the projectile? (Take g = 10 m/s²)

Core Concept

The horizontal range R of a projectile is given by R = (v₀² sin(2θ))/g, where v₀ is the initial velocity and θ is the launch angle. Substituting the given values, R = ((30)² sin(120°))/10 = (900 × 0.866)/10 = 77.94 m. However, this calculation assumes the projectile lands at the same height it was launched from. In this case, since it's launched from and lands at ground level, we can also use R = v₀x × t, where v₀x is the horizontal component of velocity and t is the total time of flight. v₀x = 30 cos(60°) = 15 m/s. The time to reach maximum height is t₁ = v₀y/g = (30 sin(60°))/10 = 2.598 s. The total time of flight is 2 × t₁ = 5.196 s. Therefore, R = 15 m/s × 5.196 s = 77.94 m. Option A is incorrect as it would be the result if we used sin(60°) instead of sin(120°). Option B is incorrect as it would be the result if we used the vertical component of velocity. Option C is the correct calculation but doesn't account for the projectile landing at ground level, which is already accounted for in the formula used.