Trajectory of a projectile

In physics, the ballistic trajectory of a projectile is the path that a thrown object will take under the action of gravity, neglecting all other forces, such as friction from air resistance, or propulsion. This article provides a list of methods for calculating the trajectory of a projectile under the influence of Earth's gravity.

Contents

Angle of reach

The "angle of reach" (not quite a scientific term) is the angle θ at which a projectile must be at in order to go a distance d, given the velocity v and distance d itself.

<math> \sin(2\theta) = \frac{9.81\,d}{v^2} </math>

Example of a baseball

<math> \sin(2\theta) = \frac{9.81\,27.4 m}{40^2} </math>


<math> \sin(2\theta) = \frac{268.794}{1600} </math>


θ = 4.866°

Time of flight

The time of flight t is the time it takes for the projectile to finish its trajectory.

<math> t = \frac{d}{v \cos\theta} </math>

Maximum distance

The maximum distance D is the distance the projectile can go at a velocity v at 45º.

<math> D = \frac {v^2}{9.81} </math>

Height at x

The "height at x" is self-explanatory, the height of the projectile y at distance x including the initial height i.

<math> y = x\tan\theta - \frac {9.81x^2}{2(v\cos\theta)^2} + i </math>

Velocity at x

The velocity at x is the current velocity s given x, θ, and initial velocity v.

<math> s^2 = (v\sin\theta)^2 - 19.6[x\tan\theta - \frac {9.81x^2}{2v\cos\theta}] = (v\sin\theta)^2 - 19.62y </math>

Distance traveled

The distance traveled by a projectile given initial angle and velocity is represented by D.

<math> D = \frac {v^2\sin(2\theta)}{9.81} </math>

Plotting the motion of a projectile

By using height at x, y, and z-deviation (the tangent of the angle of deviation from the direction the source is facing, multiplied by x), you can plot the trajectory of the projectile in three dimensions.