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Mathematics - Ordinary Differential Equations

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A ball weighing 6lb is thrown vertically downward toward the earth from a height of 1000ft with an initial velocity of 6ft/s. As it falls it is acted upon by air resistances that is numerically equal to 2/3v (in pounds), where v is the velocity (in feet per second)

a) What is the velocity and distance at the end of one minute?
b) With what velocity does the ball strike the earth?
c) At what time does the ball hit the earth?

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Solution Summary

The ordinary differential equations are examined. The velocity of the ball striking the earth is determined. A complete, neat and step-by-step solution is provided.

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W = 6 lb, m = W/g = 6/32 = 0.1875

The differential equation of the motion of the ball is therefore

m * a = Net downward force

0.1875 (dv/dt) = 6 - (2/3)v, with the initial condition v(0) = 6

0.09375 (dv/dt) = (9 - v)/3

0.09375 dv/(9 - v) = (1/3) dt

Upon integration, we get

-0.09375 ln (9 - v) = (1/3)t + ...

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