Book creator (disable)

High School Physics/Projectile motion

From Wikibooks, the open-content textbooks collection

Jump to: navigation, search

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

Relative to a flat terrain, let the initial horizontal speed be vh, and the initial vertical speed be vv. It will be shown that, the range is 2vhvv / g, and the maximum altitude is {v_v^2}/2g. The maximum range, for a given total initial speed v, is obtained when vh = vv, i.e. the initial angle is 45 degrees. This range is v2 / g, and the maximum altitude at the maximum range is a quarter of that.

Contents

[edit] Derivation

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

Let:

t \; be the time into the flight of the projectile
d_h(t) \; be the horizontal displacement at time t
d_v(t) = z \; be the vertical displacement at time t
v_h \; be the horizontal velocity (which is constant)
v_v \; be the initial vertical velocity upwards
v \; be the initial speed
v_v(t) \; be the vertical velocity at time t

Along the horizontal dimension, vh is a constant and thus by the equations of motion,

d_h(t)=v_h t \; (Equation 1)

The vertical distance, or altitude, follows the equations of motion for constant negative acceleration g:

d_v=v_vt- {{gt^2} \over 2} (Equation 2)
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 dv(t) is zero again and intercepts the ground. This occurs when dv in equation 2 is zero:

0=v_vt- {{gt^2} \over 2 }

Solving this for time t gives the time of the projectile's flight:

t={2v_v \over g } (Equation 4: "hang time" of projectile)

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

d_h(t) = { v_h t } = { v_h ({2 v_v \over g})} = { {2 v_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:

0=v_v-gt \;

Solving for t

t= { {v_v} \over g }

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

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.

[edit] Derivation in polar coordinates

In terms of angle of elevation θ and initial speed v:

v = \sqrt{ {v_h}^2 + {v_v}^2 }
v_h=v \cos \theta \;
v_v=v \sin \theta \;

Substituting into Equation 1 gives:

d_h=v_h t = vt \cos \theta \; (Equation 1a)

Substituting into Equation 2 gives:

d_v=v_vt- {{gt^2} \over 2} = ( v \sin \theta ) t - { { g t^2 } \over 2 } (Equation 2a)

Taking the derivative gives the vertical velocity:

v_v=\frac{d}{dt} d_v = \frac{d}{dt} (v \sin \theta) t - {{g t^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:

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

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

d_h=v_h t = { vt \cos \theta } =v ({{2 v \sin \theta} \over g}) \cos \theta = { { 2 v^2 \sin \theta \cos \theta } \over g}

Now using the trigonometric identity for 2sinθcosθ = sin2θ:

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

This may be solved for angle θ to give the "angle" equation to hit a target at range dh:

{ \theta } = \frac 1 2 \sin^{-1} ( { {g d_h} \over { v^2 } } ) (Equation 7: angle of projectile launch)

Note that the sine function is such that there are two solutions for θ for a given range dh. 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 t:

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

Now substitute into the vertical height equation 2a:

d_v = (v \sin \theta) t - \frac{g t^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)

[edit] 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 vh and vv such that 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 vv by solving for vh:

v_h=\sqrt{v^2-v_v^2} (Equation 8)

And substituting vh into equation 5:

d_h={ {2 v_h v_v} \over g } = { {2v_v \sqrt{v^2-v_v^2} } \over g}

The maximum dh may be determined by calculating the derivative and setting it to zero. The derivative is calculated as follows:

\frac{d}{dv_v} d_h={1 \over g} \frac{d}{dv_v} {2v_v \sqrt{v^2-v_v^2} }
={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)
={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)
={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)
={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 v:

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

Thus maximum range occurs when v2 is 2v_v^2 and this can be substituted back into equation 8:

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 vh = vv.

The actual maximum range may now be calculated by substituting vh = vv and equation 9 into equation 5:

d_h = { {2 v_h v_v} \over g } = { {2 v_v v_v} \over g } = { {2 v_v^2} \over g } = {v^2 \over g}

[edit] Maximum range in polar coordinates

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

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

Set to zero and solve for θ:

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

Now cosine is zero at \pi \over 2:

2 \theta=\frac \pi 2 (also directly clear from equation 5a, it gives the maximum possible sine value of 1)
{\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:

d_h={{ v^2 \sin 2 \theta } \over g}={{v^2 \sin 2 ( \frac{\pi}{4} ) } \over g} = {v^2 \over g}

[edit] 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:

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:

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

[edit] 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 t and substituting into equation 2:

t=\frac{d_h}{v_h} (rearrange equation 1 for t)

Substituting this into equation 2:

d_v=v_vt- \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

y=ax^2+bx+c \quad

where

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:

t=\frac{d_h}{v \cos \theta}

And this may be substituted into equation 2a:

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 tanθ = sinθ / cosθ, so:

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 y = ax2 + bx + c where y=d_v\,, x=d_h\,, a=-g/{2v^2{\cos}^2 \theta}\,, b = sinθ / cosθ and 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:

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 2 \sin \theta \cos \theta = \sin 2 \theta\,, the intersections are:

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{2 v^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.

Retrieved from "http://en.wikibooks.org/wiki/High_School_Physics/Projectile_motion"