Range of a projectile

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The path of this projectile launched from a height y0 has a range d.

The path of this projectile launched from a height y₀ has a range d.

In physics, a projectile launched with specific initial conditions in a uniform gravity field will have a predictable range. As in Trajectory of a projectile, we will use:

g: the gravitational acceleration—usually taken to be 9.81 m/s² near the Earth's surface
θ: the angle at which the projectile is launched
v: the velocity at which the projectile is launched
y₀: the initial height of the projectile
d: the total horizontal distance travelled by the projectile

When neglecting air resistance, the range of a projectile will be

$d = \frac{v \cos \theta}{g} \left( v \sin \theta + \sqrt{(v \sin \theta)^2 + 2gy_0} \right)$

If (y₀) is taken to be zero, meaning the object is being launched on flat ground, the range of the projectile will then simplify to

$d = \frac{v^2}{g} \sin 2 \theta$

[edit] Derivations

[edit] Flat Ground

First we examine the case where (y₀) is zero. The horizontal position (x(t)) of the projectile is

$x(t) = \frac{}{} v\cos \left(\theta\right) t$

In the vertical direction

$y(t) = \frac{} {} v\sin \left(\theta\right) t - \frac{1} {2} g t^2$

We are interested in the time when the projectile returns to the same height it originated at, thus

$0 = \frac{} {} v\sin \left(\theta\right) t - \frac{1} {2} g t^2$

By applying the quadratic formula

$\frac{} {}t = 0$

$t = \frac{2 v \sin \theta} {g}$

The first solution corresponds to when the projectile is first launched. The second solution is the useful one for determining the range of the projectile. Plugging this value for (t) into the horizontal equation yields

$x = \frac {2 v^2 \cos \left(\theta\right) \sin \left(\theta\right)} {g}$

Applying the trigonometric identity

$\sin(2x) = 2 \sin (x) \cos(x) \$

allows us to simplify the solution to

$d = \frac {v^2} {g} \sin 2 \theta$

Note that when (θ) is 45°, the solution becomes

$d = \frac {v^2} {g}$

[edit] Uneven Ground

Now we will allow (y₀) to be nonzero. Our equations of motion are now

$x(t) = \frac{}{} v\cos \left(\theta\right) t$

and

$y(t) = y_0 + \frac{} {} v\sin \left(\theta\right) t - \frac{1} {2} g t^2$

Once again we solve for (t) in the case where the (y) position of the projectile is at zero (since this is how we defined our starting height to begin with)

$0 = y_0 + \frac{} {} v\sin \left(\theta\right) t - \frac{1} {2} g t^2$

Again by applying the quadratic formula we find two solutions for the time. After several steps of algebraic manipulation

$t = \frac {v \sin \theta} {g} \pm \frac {\sqrt{\left(v \sin \theta\right)^2 + 2 g y_0}} {g}$

The square root must be a positive number, and since the velocity and the cosine of the launch angle can also be assumed to be positive, the solution with the greater time will occur when the positive of the plus or minus sign is used. Thus, the solution is

$t = \frac {v \sin \theta} {g} + \frac {\sqrt{\left(v \sin \theta\right)^2 + 2 g y_0}} {g}$