# High School Physics/Projectile motion

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The case of uniform gravity , disregarding drag and wind, yields a projectile motion trajectory which is a parabola. To model this, one chooses ${\displaystyle V=mgz}$, where ${\displaystyle g}$ (gee) is the so-called acceleration of gravity.

Relative to a flat terrain, let the initial horizontal speed be ${\displaystyle v_{h}}$, and the initial vertical speed be ${\displaystyle v_{v}}$. It will be shown that, the range is ${\displaystyle 2v_{h}v_{v}/g}$, and the maximum altitude is ${\displaystyle {v_{v}^{2}}/2g}$. The maximum range, for a given total initial speed ${\displaystyle v}$, is obtained when ${\displaystyle v_{h}=v_{v}}$, i.e. the initial angle is 45 degrees. This range is ${\displaystyle v^{2}/g}$, and the maximum altitude at the maximum range is a quarter of that.

## Derivation

The equations of motion may be used to calculate the characteristics of the trajectory.

Let:

${\displaystyle t\;}$ be the time into the flight of the projectile
${\displaystyle d_{h}(t)\;}$ be the horizontal displacement at time t
${\displaystyle d_{v}(t)=z\;}$ be the vertical displacement at time t
${\displaystyle v_{h}\;}$ be the horizontal velocity (which is constant)
${\displaystyle v_{v}\;}$ be the initial vertical velocity upwards
${\displaystyle v\;}$ be the initial speed
${\displaystyle v_{v}(t)\;}$ be the vertical velocity at time t

Along the horizontal dimension, ${\displaystyle v_{h}}$ is a constant and thus by the equations of motion,

${\displaystyle d_{h}(t)=v_{h}t\;}$ (Equation 1)

The vertical distance, or altitude, follows the equations of motion for constant negative acceleration ${\displaystyle g}$:

${\displaystyle d_{v}=v_{v}t-{{gt^{2}} \over 2}}$ (Equation 2)
${\displaystyle v_{v}={\frac {d}{dt}}d_{v}(t)=v_{v}-gt}$ (Equation 3: velocity equation which is the derivative of equation 2)

The range of the projectile occurs when ${\displaystyle d_{v}(t)}$ is zero again and intercepts the ground. This occurs when ${\displaystyle d_{v}}$ in equation 2 is zero:

${\displaystyle 0=v_{v}t-{{gt^{2}} \over 2}}$

Solving this for time ${\displaystyle t}$ gives the time of the projectile's flight:

${\displaystyle t={2v_{v} \over g}}$ (Equation 4: "hang time" of projectile)

The maximum range occurs when equation 4 is substituted into equation 1:

${\displaystyle d_{h}(t)={v_{h}t}={v_{h}({2v_{v} \over g})}={{2v_{h}v_{v}} \over g}}$ (Equation 5: range of projectile)

The maximum altitude for a given trajectory occurs when the vertical velocity is zero. Thus set equation 3 to zero:

${\displaystyle 0=v_{v}-gt\;}$

Solving for ${\displaystyle t}$

${\displaystyle t={{v_{v}} \over g}}$

This can be substituted into equation 2 to give the maximum altitude:

${\displaystyle d_{v}(max)=v_{v}({v_{v} \over g})-{\frac {1}{2}}g({v_{v} \over g})^{2}={{v_{v}}^{2} \over 2g}}$ (Equation 6: maximum altitude of projectile)

Thus, not surprisingly, for a given initial speed the attained altitude is highest if the initial velocity was straight up. This altitude is twice the attained altitude when the range is maximized.

## Derivation in polar coordinates

In terms of angle of elevation ${\displaystyle \theta }$ and initial speed ${\displaystyle v}$:

${\displaystyle v={\sqrt {{v_{h}}^{2}+{v_{v}}^{2}}}}$
${\displaystyle v_{h}=v\cos \theta \;}$
${\displaystyle v_{v}=v\sin \theta \;}$

Substituting into Equation 1 gives:

${\displaystyle d_{h}=v_{h}t=vt\cos \theta \;}$ (Equation 1a)

Substituting into Equation 2 gives:

${\displaystyle d_{v}=v_{v}t-{{gt^{2}} \over 2}=(v\sin \theta )t-{{gt^{2}} \over 2}}$ (Equation 2a)

Taking the derivative gives the vertical velocity:

${\displaystyle v_{v}={\frac {d}{dt}}d_{v}={\frac {d}{dt}}(v\sin \theta )t-{{gt^{2}} \over 2}=v\sin \theta -gt}$ (Equation 3a: vertical velocity)

Hang time calculated above in equation 4 may be expressed in terms of angle of elevation:

${\displaystyle t={2v_{v} \over g}={{2v\sin \theta } \over g}}$ (Equation 4a)

Equation 4a may be substituted into Equation 1a to get the horizontal distance or range:

${\displaystyle d_{h}=v_{h}t={vt\cos \theta }=v({{2v\sin \theta } \over g})\cos \theta ={{2v^{2}\sin \theta \cos \theta } \over g}}$

Now using the trigonometric identity for ${\displaystyle 2\sin \theta \cos \theta =sin2\theta }$:

${\displaystyle d_{h}={{v^{2}\sin 2\theta } \over g}}$ (Equation 5a: range of projectile)

This may be solved for angle ${\displaystyle \theta }$ to give the "angle" equation to hit a target at range ${\displaystyle d_{h}}$:

${\displaystyle {\theta }={\frac {1}{2}}\sin ^{-1}({{gd_{h}} \over {v^{2}}})}$ (Equation 7: angle of projectile launch)

Note that the sine function is such that there are two solutions for ${\displaystyle \theta }$ for a given range ${\displaystyle d_{h}}$. Physically, this corresponds to a direct shot versus a mortar shot up and over obstacles to the target.

The maximum altitude for a given range may be determined by setting the vertical velocity to zero in equation 3a and solving for ${\displaystyle t}$:

${\displaystyle 0=v\sin \theta -gt\;}$
${\displaystyle t={\frac {v\sin \theta }{g}}}$ (rearrange and solve for ${\displaystyle t}$)

Now substitute into the vertical height equation 2a:

${\displaystyle d_{v}=(v\sin \theta )t-{\frac {gt^{2}}{2}}=(v\sin \theta )({\frac {v\sin \theta }{g}})-{\frac {g({\frac {v\sin \theta }{g}})^{2}}{2}}={\frac {v^{2}{\sin }^{2}\theta }{g}}-{\frac {v^{2}{\sin }^{2}\theta }{2g}}={\frac {v^{2}{\sin }^{2}\theta }{2g}}}$ (Equation 6a: max altitude for a given launch angle)

## Maximum range

Given the above range and altitude equations, the maximum range and altitude may be determined. Both equations for the range, equations 5 and 5a may be used to determine the maximum range by setting their derivatives to zero. For equation 5, range of the projectile is a function of ${\displaystyle v_{h}}$ and ${\displaystyle v_{v}}$ such that ${\displaystyle v_{v}^{2}+v_{h}^{2}=v^{2}}$ where v is the total initial velocity and is constant. Thus, the range may be expressed as a function of ${\displaystyle v_{v}}$ by solving for ${\displaystyle v_{h}}$:

${\displaystyle v_{h}={\sqrt {v^{2}-v_{v}^{2}}}}$ (Equation 8)

And substituting ${\displaystyle v_{h}}$ into equation 5:

${\displaystyle d_{h}={{2v_{h}v_{v}} \over g}={{2v_{v}{\sqrt {v^{2}-v_{v}^{2}}}} \over g}}$

The maximum ${\displaystyle d_{h}}$ may be determined by calculating the derivative and setting it to zero. The derivative is calculated as follows:

${\displaystyle {\frac {d}{dv_{v}}}d_{h}={1 \over g}{\frac {d}{dv_{v}}}{2v_{v}{\sqrt {v^{2}-v_{v}^{2}}}}}$
${\displaystyle ={1 \over g}(2{\sqrt {v^{2}-{v_{v}}^{2}}}+2v_{v}{\frac {d}{dv_{v}}}{\sqrt {v^{2}-v_{v}^{2}}}}$ (application of product rule)
${\displaystyle ={1 \over g}(2{\sqrt {v^{2}-{v_{v}}^{2}}}+2v_{v}{\frac {d{\sqrt {v^{2}-v_{v}^{2}}}}{d{v_{v}}^{2}}}{\frac {d{v_{v}}^{2}}{dv_{v}}})}$ (application of chain rule)
${\displaystyle ={1 \over g}(2{\sqrt {v^{2}-{v_{v}}^{2}}}+2v_{v}({\frac {-1}{2{\sqrt {v^{2}-{v_{v}}^{2}}}}})2v_{v})}$ (derivative of square root)
${\displaystyle ={1 \over g}(2{\sqrt {v^{2}-{v_{v}}^{2}}}-{\frac {2{v_{v}}^{2}}{\sqrt {v^{2}-{v_{v}}^{2}}}})}$ (simplify second term)

Set to zero and solve for ${\displaystyle v}$:

${\displaystyle 0={1 \over g}(2{\sqrt {v^{2}-{v_{v}}^{2}}}-{\frac {2{v_{v}}^{2}}{\sqrt {v^{2}-{v_{v}}^{2}}}})}$
${\displaystyle {\sqrt {v^{2}-{v_{v}}^{2}}}={\frac {{v_{v}}^{2}}{\sqrt {v^{2}-{v_{v}}^{2}}}}}$
${\displaystyle v^{2}-{v_{v}}^{2}=v_{v}^{2}}$
${\displaystyle v^{2}=2v_{v}^{2}}$ (Equation 9)

Thus maximum range occurs when ${\displaystyle v^{2}}$ is ${\displaystyle 2v_{v}^{2}}$ and this can be substituted back into equation 8:

${\displaystyle v_{h}={\sqrt {v^{2}-v_{v}^{2}}}={\sqrt {v^{2}-v_{v}^{2}}}={\sqrt {2v_{v}^{2}-v_{v}^{2}}}=v_{v}}$

Thus the maximum range occurs when ${\displaystyle v_{h}=v_{v}}$.

The actual maximum range may now be calculated by substituting ${\displaystyle v_{h}=v_{v}}$ and equation 9 into equation 5:

${\displaystyle d_{h}={{2v_{h}v_{v}} \over g}={{2v_{v}v_{v}} \over g}={{2v_{v}^{2}} \over g}={v^{2} \over g}}$

## Maximum range in polar coordinates

The same conclusion may be drawn by starting with equation 5a.

${\displaystyle {\frac {d}{d\theta }}d_{h}={\frac {d}{d\theta }}{{v^{2}\sin 2\theta } \over g}}$
${\displaystyle ={\frac {v^{2}}{g}}{\frac {d}{d2\theta }}\sin 2\theta {\frac {d2\theta }{d\theta }}}$ (application of chain rule)
${\displaystyle ={\frac {v^{2}}{g}}(\cos 2\theta )2\quad }$

Set to zero and solve for ${\displaystyle \theta }$:

${\displaystyle 0={\frac {v^{2}}{g}}(\cos 2\theta )2}$
${\displaystyle 0=\cos 2\theta \quad }$

Now cosine is zero at ${\displaystyle \pi \over 2}$:

${\displaystyle 2\theta ={\frac {\pi }{2}}}$ (also directly clear from equation 5a, it gives the maximum possible sine value of 1)
${\displaystyle {\theta }_{\max }={\frac {\pi }{4}}}$ radians

Thus the maximum range occurs when the angle is 45 degrees.

The actual maximum range may now be calculated by substituting 45 degrees into equation 5a:

${\displaystyle d_{h}={{v^{2}\sin 2\theta } \over g}={{v^{2}\sin 2({\frac {\pi }{4}})} \over g}={v^{2} \over g}}$

## Maximum altitude at maximum range

Equations 6 and 6a may be used to calculate the maximum altitude at the maximum range. Equation 9 may be substituted into equation 6:

${\displaystyle d_{v}={\frac {v_{v}^{2}}{2g}}={\frac {({\frac {v^{2}}{2}})}{2g}}={\frac {v^{2}}{4g}}}$

Likewise 45 degrees may be substituted into equation 6a:

${\displaystyle d_{v}={\frac {v^{2}{\sin }^{2}\theta }{2g}}={\frac {v^{2}({\frac {1}{\sqrt {2}}})^{2}}{2g}}={\frac {v^{2}}{4g}}}$

## As a parabola

Equations 1 and 2 are parametric equations that describe a parabola. They may be rearranged into the more familiar quadratic form by solving equation 1 for ${\displaystyle t}$ and substituting into equation 2:

${\displaystyle t={\frac {d_{h}}{v_{h}}}}$ (rearrange equation 1 for ${\displaystyle t}$)

Substituting this into equation 2:

${\displaystyle d_{v}=v_{v}t-{\frac {gt^{2}}{2}}=v_{v}({\frac {d_{h}}{v_{h}}})-{\frac {g({\frac {d_{h}}{v_{h}}})^{2}}{2}}=-{\frac {g}{2{v_{h}}^{2}}}(d_{h})^{2}+{\frac {v_{v}}{v_{h}}}(d_{h})}$

This is now in the form

${\displaystyle y=ax^{2}+bx+c\quad }$

where

${\displaystyle y=d_{v},x=d_{h},a=-g/{2{v_{h}}^{2}},b=v_{v}/v_{h},c=0}$.

This is the form of a parabola and thus the trajectory is a parabola.

Likewise equations 1a and 2a can be rearranged into quadratic form. Equation 1a may be rearranged to:

${\displaystyle t={\frac {d_{h}}{v\cos \theta }}}$

And this may be substituted into equation 2a:

${\displaystyle d_{v}=vt\sin \theta -{\frac {gt^{2}}{2}}=v\sin \theta {\frac {d_{h}}{v\cos \theta }}-{\frac {g{({\frac {d_{h}}{v\cos \theta }})}^{2}}{2}}}$

Now ${\displaystyle \tan \theta =\sin \theta /\cos \theta }$, so:

${\displaystyle d_{v}=-{\frac {g}{2v^{2}{\cos }^{2}\theta }}d_{h}^{2}+d_{h}\tan \theta }$ (Equation 10)

This is again now in the form ${\displaystyle y=ax^{2}+bx+c}$ where ${\displaystyle y=d_{v}\,}$, ${\displaystyle x=d_{h}\,}$, ${\displaystyle a=-g/{2v^{2}{\cos }^{2}\theta }\,}$, ${\displaystyle b=\sin \theta /\cos \theta }$ and ${\displaystyle c=0\,}$ demonstrating that this is a parabola.

The quadratic formula gives the location of the intersection of the parabola and the x-axis. This is where the projectile trajectory starts and ends and thus may be used directly to calculate the range. In terms of rectilinear coordinate systems:

${\displaystyle d_{h}={\frac {-b\pm {\sqrt {b^{2}-4ac\ }}}{2a}}={\frac {-(v_{v}/v_{h})\pm {\sqrt {(v_{v}/v_{h})^{2}-0}}}{2(-g/2v_{h}^{2})}}={\frac {-v_{v}/v_{h}\pm v_{v}/v_{h}}{-g/v_{h}^{2}}}=0,{\frac {2{v_{v}}{v_{h}}}{g}}\,}$

This is the same result as equation 5 above.

In polar coordinates and using the trigonometric identity ${\displaystyle 2\sin \theta \cos \theta =\sin 2\theta \,}$, the intersections are:

${\displaystyle d_{h}={\frac {-b\pm {\sqrt {b^{2}-4ac\ }}}{2a}}={\frac {-(\sin \theta /\cos \theta )\pm {\sqrt {(\sin \theta /\cos \theta )^{2}-0}}}{2(-g/2v^{2}\cos ^{2}\theta )}}=0,{\frac {2\sin \theta /\cos \theta }{g/v^{2}\cos ^{2}\theta }}=0,{\frac {2v^{2}\sin \theta \cos \theta }{g}}=0,{\frac {v^{2}\sin 2\theta }{g}}\,}$

This is the same result as in equation 5a above.

Similarly, the vertex of the parabola is the maximum altitude for a given range.

## Computer simulation

NCLab provides an interactive graphical module for projective motion with and without air friction. The Python source code of the simulation can be freely view and copied. For the case with air friction, Runge-Kutta methods of orders 1, 2 and 4 are used to solve the underlying ordinary differential equations.