A projectile is launched with an initial velocity of 60.0 m/s at an angle of 30.0° above the horizontal. How far does it travel?

To determine the horizontal distance, use the range equation.

Range = v^2/g * sin 2 θ

Range = 60^2/9.8 * sin 60 = 318.131781 meters

You can use this equation when the initial and final heights of the projectile are equal.

Physics projectile problem assumes no air friction.

Vo = 60.0 m/s

Θ = 30.0”

Voy = Vo(sin Θ) = 60.0(sin 30.0°) = 30.0 m/s

Vox = Vo(cos Θ) = 60.0(cos 30.0°) = 52.0 m/s

time to reach max height = t = Voy/g = 30.0/9.81 = 3.058104 s

total time of flight = 2t = 6.116208 s

range of flight = R = Vox(2t) = 52.0(6.116208) = 318.0428 ≈ 318 m ANS

THANK YOU SO MUCH!!! Now at least I know how to do that!