Level-Ground Projectile Range

R = \frac{v _{0}^{2} sin 2 θ}{g}

For launch and landing at the same elevation with no drag, range is maximum at θ = 45° (sin 2θ = 1). Complementary angles (e.g. 30° and 60°) give the same R. Horizontal acceleration is zero.

$R$ (m): Horizontal range
$v_{0}$ (m/s): Launch speed
$θ$ (deg): Launch angle above horizontal
$g$ (m/s²): Gravity

Step-by-step solved example

v0 = 20 m/s, θ = 30°, g = 9.81 m/s², level ground. Find R.

Level-ground projectile is a parabola; R = v0² sin 2θ / g.

1. Double angle
sin 2θ = sin 60° = 0.8660.
$sin 6 0^{\circ} = 0.8660$
2. Range
R = (400)(0.8660)/9.81 = 35.3 m.
$R = 35.3 m$

Answer: 35.3 m

10 practice questions

0/10 correct

1.v0 = 10 m/s, θ = 45°, g = 10 m/s². R =

2.Maximum range on level ground occurs at

3.Complementary launch angles 30° and 60°

4.If v0 doubles, level-ground R

5.Horizontal acceleration of a projectile (no drag) is

6.Time of flight on level ground is

7.v0 = 30 m/s, θ = 15°, g = 10 m/s². R =

8.The range formula R = v0² sin2θ / g assumes

9.v0 = 14 m/s, θ = 45°, g = 9.8 m/s². R =

10.At θ = 90° (straight up) the range is