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Encyclopedia > Range of a projectile

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: The first few hydrogen atom electron orbitals shown as cross-sections with color-coded probability density. ... A projectile is any object sent through space by the application of a force. ... In mathematics, boundary conditions are imposed on the solutions of ordinary differential equations and partial differential equations, to fit the solutions to the actual problem. ... Gravity is a force of attraction that acts between bodies that have mass. ... 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. ...

g: the acceleration due to gravity—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 For other uses, see g force. ...

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

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

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

Derivations

Flat Ground

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

$x(t) = frac{}{} vcos left(thetaright) t$

In the vertical direction

$y(t) = frac{} {} vsin left(thetaright) t - frac{1} {2} g t^2$

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

$0 = frac{} {} vsin left(thetaright) t - frac{1} {2} g t^2$

By applying the quadratic formula In mathematics, a quadratic equation is a polynomial equation of the second degree. ...

$frac{} {}t = 0$

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

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

$x = frac {2 v^2 cos left(thetaright) sin left(thetaright)} {g}$

Applying the trigonometric identity In mathematics, trigonometric identities are equations involving trigonometric functions that are true for all values of the occurring variables. ...

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

allows us to simplify the solution to

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

Note that when theta is 45°, the solution becomes

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

Uneven Ground

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

$x(t) = frac{}{} vcos left(thetaright) t$

and

$y(t) = y_0 + frac{} {} vsin left(thetaright) t - frac{1} {2} g t^2$

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)

$0 = y_0 + frac{} {} vsin left(thetaright) t - frac{1} {2} g t^2$

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

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

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

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

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