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# Projectile Motion - Christian Brothers University

By Stanley Black,2014-03-17 22:37
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Projectile Motion - Christian Brothers UniversityProjec

Projectile Motion

1. Without air resistance (so 2-D): F = -mgy

Since the force does not depend on x, y, or z, it is easier to use the rectangular system. And since the force does not depend on x, y, or z, we can separate the 3-D motion into three 1-D equations, and so use the techniques of Chapter 2.

1a) x equation: ;F = 0 = ma leads to a=0 which leads to: xxx

v = v and x = x + vt (Eq.1) xxooxo

1b) y equation: ;F = -mg = ma leads to a=-g which leads to: yyy

2v = v - gt and y = y + vt - (1/2)gt (Eq.2) yyooyo

1c) z equation: ;F = 0 = ma , and since v=0, there is and will be no motion in the zzzo

z-direction; this makes the problem just 2-D.

1d) y(x) trajectory: These equations for x(t) and y(t) are the parametric equations for the trajectory. But to actually find where the x will be at any particular y (such as where the object will hit the ground), we have to solve y(t) for t and then use this t in the x(t). Hence it might be nice to express the trajectory as y(x), which means we must eliminate t from these equations. Solving for t(x) from Eq. 1, we have:

t = (x-x)/v . (Eq. 1a) oxo

We now put this expression in for t in Eq. 2:

22y = y + vt - (1/2)gt = y + v[(x-x)/v] - (1/2)g[(x-x)/v] . oyooyooxooxo

It will be simpler if we let x=0 and y=0. We can always go back and replace x with (x-oo

x) and replace y with (y-y) if these are not initially zero. oo

22y(x) = (v/v)x - (1/2)gx /v (Eq. 2a) yoxoxo

This is a parabola, so we should be able to put this in standard parabolic form

2(x-x) = k(y-y) . pp

2We start by getting the coefficient of the x term in Eq. 2a to be 1:

22-2vy/g = -2vvx/g + x . xoxoyo

2We now complete the square by adding a constant term: [vv/g] to both sides: xoyo

222-2vy/g + [vv/g] = [x - vv/g] xoxoyoxoyo

22222[x - vv/g] = [vv/g] - 2vy/g = (-2v/g)[y - v/2g] xoyoxoyoxoxoyo

222[x - vv/g] = (-2v/g)[y - v/2g] . (Eq. 3) xoyoxoyo

This last form, Eq. 3, is the standard form for a parabola. For y to be at a maximum, dy/dx = 0 which, using Eq. 2a, gives:

222d[(v/v)x - (1/2)gx /v]/dx = (v/v) - gx/v = 0 yoxoxoyoxoxo

2which means that x = vv/g for y at its peak. But this means that y = v/2g . peakxoyopeakyoThus Eq. 3 becomes:

22[x - x] = (-2v/g)[y - y] . (Eq. 3a) peakxopeak

1e) Range: This form (Eq. 3) also allows an easy calculation of the range of the

trajectory (that is, the value of x when y=0). [Remember that by substituting y=0 into Eq.

3, we are also assuming y=0 as well.] o

22222222[x - x] = (-2v/g)[0 - y] = 2vy/g = 2v[v/2g]/g = vv/g rangepeakxopeakxopeakxoyoxoyo

x = x + vv/g = vv/g + vv/g = 2 vv/g = x . (Eq. 4) rangepeakxoyoxoyoxoyoxoyorange

2. With air resistance, but no wind (so still 2-D): F = -mgy - bv Since the force does not depend on x, y, or z, and since the force depends linearly on v (that is, F depends on v but not v or v, etc), it is easier to use the rectangular system xxyz

and separate the 3-D motion into three 1-D equations. . [Note: we could use a resistive force 2that is proportional to v which would be more realistic. It could also be broken into rectangular 222components, e.g., F = -bvv . However, this adds a complication since v = SQRT[v + v + v]. AR-yyxyz

2a) x equation: ;F = -bv = ma = m dv/dt . xxxx

Using the form: -bv = m dv/dt gives the expression: xx

(-b/m)dt = dv/v . xx

Integrating both sides gives:

tvx (-b/m) dt = dv/v 0vxoxx

which integrates to

-bt/m = ln[v/v] or t = (-b/m) ln[v/v] , xxoxxo

-bt/mor solving for v(t): v(t) = v e . xxxo

-bt/mTo get x(t), we use dx/dt = v(t), or dx = v e dt . Integrating gives: xxo

xtt-bt/m dx = v(t) dt = v e dt , or xo0x0xo

-bt/m0-bt/mx - x = (-m/b)v e - (-m/b)v e = (mv/b)(1 - e) . (Eq. 5) oxoxoxo

Anticipating the need to put t(x) into the y(t) equation to get y(x), we try to invert Eq. 5 to

get t(x). As before, we will assume x=0 since we will always be able to replace x with o

(x-x) if that is not true. o

-bt/mx = (mv/b)(1 - e) (Eq. 5) becomes xo

-bt/m(bx/mv) = (1 - e) (Eq. 5a), or xo

-bt/me = 1 - (bx/mv), or t = (-m/b) ln[1 - (bx/mv)] . (Eq. 5b) xoxo

2b) y equation: ;F = -mg -bv = ma = m dv/dt . yyyy

Using the form: -mg -bv = m dv/dt gives the expression: yy

(-b/m)dt = dv/(v+mg/b) . yy

Integrating both sides gives:

tvy (-b/m) dt = dv/(v+mg/b) . 0vyoyy

If we let u = (v+mg/b), then du = dv and we have: yy

vy+mg/b-bt/m = du/u = ln[(v+mg/b)/(v+mg/b)] . vyo+mg/byyo

We now have essentially t(v). Inverting to get v(t): yy

-bt/m(v+mg/b)/(v+mg/b) = e , or yyo

-bt/m(v+mg/b) = (v+mg/b) e , or yyo

-bt/mv = -(mg/b) + (v+mg/b) e , or yyo

-bt/m-bt/mv = ve - (mg/b)(1 - e) . (Eq. 6) yyo

-bt/m-bt/mTo get y(t), we use dy/dt = v(t) = ve - (mg/b)(1 - e) . Integrating this gives: yyo

yt-bt/m-bt/m dy = [ve - (mg/b)(1 - e)] dt , yooyo

which upon integrating term by term gives:

-bt/m -bt/m y - y = -(mv/b)(e- 1) - (mg/b)t -(m/b)(mg/b)(e- 1) . oyo

As before, we will assume y=0 since we will always be able to replace y with (y-y) if oo

that is not true. Upon a slight re-arrangement of terms, we have:

22-bt/m y(t) = - (mg/b)t + (mg/b + mv/b)(1 - e) . Eq. (7) yo

Before we proceed to try to get y(x), let's look at the z component.

2c) z equation: ;F = -bv = ma , and since v=0, there is and will be no motion in zzzzo

the z-direction; this makes the problem just 2-D.

2d) y(x) trajectory: These equations for x(t) and y(t) are the parametric equations for the trajectory. But to actually find where the x will be at any particular y (such as where

the object will hit the ground), we have to solve y(t) for t and then use this t in the x(t).

Hence it might be nice to express the trajectory as y(x), which means we must eliminate t -bt/mfrom these equations. We use Eq. 5(b) for t and Eq. 5(a) for (1 - e) in Eq. (7) to get:

22y(t) = - (mg/b) (-m/b) ln[1 - (bx/mv)] + (mg/b + mv/b) (bx/mv) , or xoyoxo

22y(x) = (mg/b) ln[1 - (bx/mv)] + (mg/bv + v/v) x . Eq. (8) xoxoyoxo

It is hard to compare the result (Eq. 8) for y(x) with the result for no air resistance (Eq. 3).

2e) special case: small air resistance (b is small)

However, we can see if Eq. 8 reduces to Eq. 3 in the case of b=0, and then see the small corrections to Eq. 3 that Eq. 8 predicts. When b goes to zero, it looks like the factor in front of the ln blows up. However, the ln goes to zero. How do we see what really happens? To do this, we can expand the ln in a power series: 234ln(1-u) = 0 - u - u/2 - u/3 u/4 + ... , so as long as u = (bx/mv) << 1, we have: xo

22234y(x) = (mg/b)[0 - (bx/mv) - (bx/mv)/2 - (bx/mv)/3 - (bx/mv)/4 - ...) + (mgx/bv) + (v/v)x xoxoxoxoxoyoxo

23242234= 0 - mgx/bv - gx/2v - bgx/3mv - bgx/4mv ... + (mgx/bv) + (v/v)x xoxoxoxoxoyoxo

23242234y(x) = (v/v)x- gx/2v- bgx/3mv - bgx/4mv - ... (Eq. 9) yoxoxoxoxo

22As b goes to zero, y(x) becomes simply y = (v/v)x- gx/2v , which is the same as yoxoxo

we obtained for no air resistance (Eq. 2a). For a small b [such that (bx/mv) << 1], we xo

33have the correction from the no air resistance result of: - bgx/3mv if we neglect terms xo

2of order b and higher. Note that this correction makes y smaller with air resistance than without, as expected.

2f) range correction for small air resistance:

To solve for the range with small air resistance, we start with Eq. (8):

22y(x) = (mg/b) ln[1 - (bx/mv)] + (mg/bv + v/v) x . Eq. (8) xoxoyoxo

With y=0 (and also assuming y=0) we get for x : orange

220 = (mg/b) ln[1 - (bx /mv)] + (mg/bv + v/v) x . rangexoxoyoxorange

This appears tough to solve for x since it appears in both the ln term and in the linear range

term! But for small air resistance [(bx /mv)<<1], we can expand the ln term like we rangexo

did to get Eq. 9 above, with y being replace by zero and x by x : range

242223340 = (v/v)x- gx/2v- bgx/3mv - bgx/4mv - ... (Eq. 10) yoxorangerangexorangexoxo

First, we look at the case where b=0 (no air resistance). Eq. 10 then gives

220 = (v/v)x- gx/2v , or yoxorangerangexo

x = 2vv/g rangexoyo

which matches the results we obtained for no air resistance! Note that if y0, the o

quadratic equation must be used to find x. range

To solve for x with the first b term, the quadratic equation must be employed. range

If y0, the equation for a cubic must be employed - unless we employ the method of o

successive approximations we employed earlier in the semester. In this method it is important to keep track of levels of approximation. To keep the algebra down, we will assume y=0. o

(0) For the zero order term, we use the results for b=0: x = 2vv/g . To rangexoyo

(0)solve for our result through first order, we substitute x into the first order b term of range

2Eq. 10, ignore all terms of order b and higher, and then solve the quadratic equation for

(1)x : range

23(1) (1)2 (0)30 = (v/v)x- g[x]/2v- bg[x]/3mv yoxorangerangexorangexo

(0)Substituting x = 2vv/g in the above equation gives: rangexoyo

232(1)(1)3(g/2v)[x] - (v/v)x + (bg/3mv)( 2vv/g) = 0 , xorangeyoxorangexoxoyo

or simplifying the last term:

222(1)(1)3(g/2v)[x] - (v/v)x + (8vb/3mg) = 0 . xorangeyoxorangeyo

We now can use the quadratic formula to get:

221/2(1)232x = {+(v/v) [(v/v) - 4(g/2v)(8vb/3mg)] }/ 2(g/2v) rangeyoxoyoxoxoyoxo

Simplifying each term gives:

21/2(1)322x = {+(v/v) [(v/v) - (32vb/6vmg)] }/ (g/v) rangeyoxoyoxoyoxoxo

Note that a 1/v can be taken out of both the numerator and the denominator: xo

21/2(1)3x = {+(v) [(v) - (32vb/6mg)]}/ (g/v) . rangeyoyoyoxo

A v can be taken out of each term in the numerator and combined with the denominator yo

to give:

1/2(1)x = (vv/g) {+1 [1 - (16vb/3mg)] }. rangexoyoyo

(1)(0)If b=0, then x = 2vv/g = x as it should. This is still an awkward formula, rangexoyorange

1/2and since b is small, we can use the expansion: [1-u] 1 - u/2 to get:

2(1)2x = (vv/g) {+1 [1 - (8vb/3mg)] } = (2vv/g) - 8vvb/3mg . rangexoyoyoxoyoxoyo

The first term is the zero order range, and thus the first order correction to the range due

22to a small air resistance is the second term: - 8vvb/3mg . Note that this term is xoyo

negative since air resistance will make the range smaller.

It is interesting to see this correction as a fraction of the total range:

(1)(0)x = (2vv/g) (1 - 4bv/3mg) = x (1 - 4bv/3mg) . (Eq.11) rangexoyoyorangeyo

3To get the next higher order correction, we ignore all terms of order b and higher,

(0)2(1)(2)substitute x into the term with b, substitute x into the term with b, and substitute x

(2)into the term with no b, and then solve for x. This is one choice for problem #20.

The table below shows the results of using m=1 kg; v=40 m/s; v=60 m/s; and various xoyo

values of b. The numerical was run on an excel worksheet with dt=0.05 seconds. The

analytical result was obtained by graphing y versus x and estimating where x is when y=0.

(0) (1) b xbx/mv1st order xResult of num. Result of rangerangexo range (Nt-s/m) (m)(should be correction (m) (m)calc. (m) graphing exact

<1)equation (m)

0.00 489.8 0.00 0.0 489.8 489.8 489.8

0.001 489.8 0.012 4.0 485.8 485.8 485.8

0.01 489.8 0.11 40.0 449.8 452.3 452.5

0.05 489.8 0.362 199.9 289.88 342.8 343.1

0.10 489.8 0.22 399.83 89.96 259.0 264.4

The table below shows the results for a baseball using m=0.145 kg; v = 100 mph = 44.64 oom/s at an initial angle of 45. The numerical calculation was run on an excel worksheet with dt=0.05 seconds. A better model for air resistance for a baseball going 100 mph 2would be F = bv, but as indicated above, this is a hard differential equation to solve. AR2

We can use our present solution, however, to get a fairly good approximation if for the

“b” value in the F = bv if we use b = bv , where the value for b = ?CAρ where C = AR2o220.3, A = cross sectional area = πr , and ρ = density of air. For a baseball, r = 3.66 cm, 3-4 and at sea level ρ = 1.2 kg/m . For these values, b = 7.6 x 10, and b = .034 . 2

(0) b xResult of Result of Result of num. range (Nt-s/m) with b=0num. calc. graphing exact calc. using 2using F=bvequation F=bv 2

0.034 203.3 m 95.6 m 96.2 m 113.7 m

667 ft 314 ft 315 ft 373 ft

Note that the result using b=bv gives a smaller range this is due to the fact that v 2o

slows down and is less than v, so the b value is too high. The result above for the o

baseball does NOT include effects due to the spinning of the ball.

2g) special case: range correction for large b

Recall the expression for y(x) which was exact (in this model):

22y(x) = (mg/b) ln[1 - (bx/mv)] + (mg/bv + v/v) x . Eq. (8) xoxoyoxo

To make this into the range equation, we replace y(x) with 0 (assuming also that y=0), oand x with x: range

220 = (mg/b) ln[1 - (bx/mv)] + (mg/bv + v/v) x . rangexoxoyoxorangeSince the argument of the ln must remain greater than 0, (bx/mv) < 1, or bx < rangexorangemv. If b is large, then we can assume that bx = mv(1-) where is small. Using xorangexo

this, we have (bx/mv) = (1-), and also x= (mv/b)(1-). Making these rangexorange xo

substitutions for x in terms of into the range equation gives: range

220 = (mg/b) ln[1 - (1-)] + (mg/bv + v/v) (mv/b)(1-) . xoyoxoxo

Expanding and simplifying, we have:

2222220 = (mg/b) ln[)] + mg/b + mv/b - mg/b - mv/b yoyoPlacing all the terms with 's on one side gives:

222222(mg/b) ln[)]- mg/b - mv/b = -mg/b - mv/b . yoyo

Under the assumption that is small, ln() >> , so we can neglect the terms linear in :

2222(mg/b) ln[)] = -mg/b - mv/b yo

22-ln[)] = 1 + (mv/b)/ (mg/b) = 1 + (bv/mg) = ln(1/) yoyo

[1 + (bvyo/mg)]1/ = e

And since (bx/mv) = (1-), we have = 1 - bx/mv : rangexorangexo

-[1 + (bvyo/mg)] = e = 1 - bx/mv rangexo

-[1 + (bvyo/mg)]x = (mv/b) [1 - e ]. (Eq. 12) rangexo

If the initial air resistance (bv) is much greater than the gravitational force (mg), then yo

the exponential term is negligible and the range will simply be:

x = (mv/b) (Eq. 13) rangexo

If we use the same values we had earlier for small air resistance: m=1 kg; v=40 m/s; xov=60 m/s; and we set b=1, we get: yo

(from Eq. 12): x = 39.97 m with =1 - bx/mv = .00075 rangerangexo

(from Eq. 13): x = 40.00 m. range

(from a numerical approximation): x = 39.56 m . range

If we use the above values and set b=0.1, we get:

(from Eq. 12): x = 320.22 m with =1 - bx/mv = .80 rangerangexo

(from Eq. 13): x = 400.00 m. range

(from a numerical approximation): x = 259 m . range

3. With air resistance and with wind:

How do we include the wind in our equations? Obviously, the wind produces a force, but how do we account for that force in Newton's Second Law? The wind is simply moving air, and so air resistance should come into play. Therefore, we simply need to add a term: +bw where b is the air resistance coefficient, and w is the wind (speed and direction).

3a) x component: ;F = -bv + bw = m dv/dt xxxx

(-b/m)dt = dv/(v-w) xxx

tvx (-b/m) dt = dv/(v-w) t=0vxoxxx

Let u = (v-w), du=dv : xxx

vx-wx-bt/m = du/u = ln[(v-w)/(v-w)] vxo-wxxxxox

-bt/m(v-w) = (v-w) e xxxox

-bt/mv(t) = w + (v-w) e xxxox

To find x(t), we use: dx(t)/dt = v(t), or x

xtt-bt/m dx = v(t) dt = [w + (v-w) e] dt xot=0xt=0xxox

-bt/m0x(t) - x = wt + (-m/b)(v-w) (e - e) oxxox

-bt/mx(t) = x + wt + (m/b)(v-w) (1 - e) (Eq. 14) oxxox

3b) y component:

If we do not consider any vertical wind, the equation for y(t) is the same as it was before, and so we get:

22-bt/m y(t) = - (mg/b)t + (mg/b + mv/b)(1 - e) . Eq. (7) yo

3c) z component:

The same situation holds for the z component as held with the x component, the only difference being that v=0. Therefore we have: zo

-bt/m-bt/mz(t) = z + wt + (m/b)(v-w) (1 - e) = z + wt + (m/b)(-w) (1 - e) . (Eq.15) ozzozozz

The term wt in Eq. 14 makes getting y(x) much harder than it was previously, because x

we no longer have the simple results relating t to x in Eqs. 5a and 5b:

-bt/m(1 - e) = (bx/mv) (Eq. 5a), and t = (-m/b) ln[1 - (bx/mv)] . (Eq. 5b) xoxo

3d) special case: purely cross wind

If w=0, we have the special case of a cross wind. In this case, Eqs. 5a and 5b still hold, x

and we get exactly the same y(x) we had with no wind. The only effect of the cross wind is to blow the object off to the side (z-direction). We can now use Eqs. 5a and 5b in Eq. 15 to get z(x). Then knowing x, we can get z . Again, like before, we will rangedrift

incorporate z into z: o

-bt/mz(t) = wt - (m/b)(w) (1 - e) ( Eq. 15) zz

z(x) = w{(-m/b) ln[1 - (bx/mv)]} - (mw/b)(bx/mv) . zxozxo

3e) special case: purely cross wind with small b

2If (bx/mv) is small, then we can expand the ln in a power series: ln(1-u) = -u -u/2 - xo

34u/3 u/4 - ... . Therefore, z(x) becomes

234z(x) = -(mw/b)[ -(bx/mv) - (bx/mv)/2 - (bx/mv)/3 - (bx/mv) - ...] - (mw/b)(bx/mv) . zxoxoxoxozxo

Notice that the first term in the series cancels the last term. Therefore, we have:

223223z(x) = wbx/2mv + wbx/3mv + ... (Eq. 16) zxozxo

To get the total drift, we now substitute x . If we want the drift to first order in b, we range

only need to substitute the zero-order range into the first term of Eq. 16. If we want the drift to second order in b, we need to substitute the first order range into the first term of Eq. 16 and substitute the zero order range into the second term. Here we will only look

(0)for the first order drift. Recall that the zero order range is: x = 2vv/g . rangexoyo

222(1)222z = wbx/2mv = wb[2vv/g]/2mv = 2wbv/mg (Eq. 17) driftzxozxoyoxozyo

As we know, the zero order drift is zero. To get some idea of the size of this drift, we will use the values we used earlier for a baseball: 2m = .145 kg; v = 44.64 m/s = 100 mph; g = 9.8 m/s , o22b = b*v = .034 Nt*s/m from b = .00076 Nt*s/m : 2o2

w Analytical Numerical First order Numerical z2 using bv2

(m/s) (m) (m) (m) (m)

10 23.0 22.8 48.4 17.8

Note that in the above case, the first order correction is not really close to the other values. If we look at the conditions for this first order condition to be good, (bx/mv) must be xo

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