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Ordinary Differential Equations
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with initial conditions as in (a). m = 5; g = 32.16; solution = DSolve[{m h''[t] m g h'[t], h'[0] 0, h[0] 1000}, h[t], t]; height2[t_]= solution[[1, 1, 2]]
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0.2 t( 804. + 1804. 0.2 t 160.8 0.2 t t)
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FindRoot[height2[t] 0, {t, 10}] {t 10.6213} It now takes 10.6213 sec to reach the ground. PlotLegends`
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Plot[{height1[t], height2[t]}, {t, 0, 11}, PlotStyle {Thick,Thin}, PlotRange {0,1000}, AxesLabel {"Time","Height"}, PlotLegend {"Without air resistance","Air resistance included"}, LegendSize {1, .5}]
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Height 1000 800 600 400 200 Without air resistance 0 Air resistance included 2 4 6 8 10 Time
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11.7 A baseball is hit with a velocity of 100 ft/sec at an angle of 30 with the horizontal. The height of the bat is 3 ft above the ground. Neglecting air and wind resistance, will it clear a 35-ft-high fence located 200 ft from home plate (Assume g = 32.16 ft/sec2.)
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SOLUTION
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g = 32.16; h = 3; = 30 Degree; v0 = 100; solution = DSolve[{x''[t] 0, y''[t] g, x'[0] v0 Cos[ ], y'[0] v0 Sin[ ], x[0] 0, y[0] h}, {x[t], y[t]}, t]; xsolution[t_]= solution[[1, 1, 2]] 50 3 t ysolution[t_]= solution[[1, 2, 2]] 3 + 50 t 16.08 t2 ParametricPlot[{xsolution[t], ysolution[t]}, {t, 0, 3.2}, AxesLabel {"x", "y"}]
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y 40 30 20 10 50 100 150 200 250 x
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Ordinary Differential Equations
From the graph it is questionable whether y 35 when x = 200, so we compute precisely when the ball reaches the fence and then calculate its height at that instant. Solve[xsolution[t] 200] t 4 3 ysolution[t] /.% {32.7101} Since the height of the ball is less than 35 ft, the ball will not clear the fence.
11.8 At what angle should the ball in the previous problem be hit so that it goes over the fence
SOLUTION
First we want to get a relationship between y and q. Clear[ ] g = 32; h = 3; v0 = 100; solution = DSolve[{x''[t] 0, y''[t] g, x'[0] v0 Cos[ ], y'[0] v0 Sin[ ], x[0] 0, y[0] h}, {x[t], y[t]}, t]
{{x[t] 100 t Cos[ ],y[t] 3 16 t
horiz[t_]= solution[[1, 1, 2]]; vert[t_]= solution[[1, 2, 2]] 3 16 t2 + 100 t Sin[ ] temp = Solve[horiz[t] 200, t]
{{t 2 Sec[ ]}}
+ 100 t Sin[ ] } }
Solve for t as a function of q. Define a function representing the height as a function of . Find which gives a height of 35 ft. The value of is expressed in degrees.
height[p_]= vert[t] /. temp
64 Sec[ ] + 200
Tan[ ]}
NSolve[height[ Degree] 35, p]
Solve ifun : Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. This warning may be safely disregarded.
{{ 149.216}, { 111.694}, { 30.7838}, { 68.3065}} The negative values of may be disregarded. The ball will go over the fence if lies between 30.6357 and 68.4546 . We conclude by sketching these two trajectories. The vertical line represents the 35-ft fence located 200 ft from home plate. = 30.6357 Degree; horiz[t_]= solution[[1, 1, 2]]; vert[t_]= solution[[1, 2, 2]]; graph1 = ParametricPlot[{horiz[t], vert[t]}, {t, 0, 6}] ; = 68.4546 Degree; horiz[t_]= solution[[1, 1, 2]]; vert[t_]= solution[[1, 2, 2]]; graph2 = ParametricPlot[{horiz[t], vert[t]}, {t, 0, 6}] ; graph3 = Graphics[Line[{{200, 0}, {200, 35}}]] ;
Ordinary Differential Equations
Show[graph1, graph2, graph3, PlotRange { 50, 150}, AxesLabel {"x", "y"}]
y 150
50 50
11.9 A culture of microorganisms grows at a rate proportional to the amount present at any given time. If there are 500 bacteria present after one day and 1200 after two days, how many bacteria will be present after four days
SOLUTION
The differential equation described by this situation is dN = kN , where N is the number of bacteria present dt in the culture and k is a constant to be determined by the given information. The initial condition is N = 500 when t = 1. solution = DSolve[{n'[t] k n[t], n[1] 500}, n[t], t]
{{n[t] 500
k+kt
population[t_]= solution[[1, 1, 2]]; Solve[population[2] 1200, k]
Solve ifun : Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.
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