Question

3. (a) An outfielder fields a baseball 280 ft away from home plate and throws it directly to the catcher with an initial velocity of 100 ft/s. Assume that the velocity v(t) of the ball after t seconds satisfies the di↵erential equation dv dt = 1 10 v because of air resistance. How long does it take for the ball to reach home plate? (Ignore any vertical motion of the ball.) (Instructor’s hint: Recall that a di↵erential equation of the form dv/dt = kv has solution v(t) = v(0)ekt.) (b) The manager of the team wonders whether the ball will reach home plate sooner if it is relayed by an infielder. The shortstop can position himself directly between the outfielder and home plate, catch the ball thrown by the infielder, turn, and throw the ball to the catcher with an initial velocity of 105 ft/s. The manager clocks the relay time of the shortstop (catching, turning, throwing) at half a second. How far from home plate should the shortstop position himself to minimize the total time for the ball to reach home plate? Should the manager encourage a direct throw or a relayed throw? What if the shortstop can throw at 115 ft/s? (Instructor’s hint: Let x represent the distance between the shortstop and home plate, then find an expression for the time it takes that ball to reach home plate as a function of x. It is also helpful to use a variable w to represent the shortstop’s throwing velocity, since you can then substitute the di↵erent given values in place of w.) (c) For what throwing velocity of the shortstop does a relayed throw take the same time as a direct throw

Please answer part b

Answer 1

Okay, Here is the solution to Part (b)

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(b)

Let 'x' be the distance from home plate to short stop, then distance from fielder to shortstop is 280 - x.

using the hint given in part (a)

we have

1 - e^{-t / 10} = (280 - x) / 1000

solve for t, we get

- e^{-t / 10} = (280 - x) / 1000 - (1)

take 'ln' on both sides, we have

t = -10 ln ((720 + x) / 1000)

in same way, we find time t' if ball is thrown at relayed velocity. Let's call it u.

so,

v = ue^-t/10

so, we have

10u ( 1 - e^{-t' / 10}) = x

again solve for t', we get

t' = - 10 ln ((10u - x) / 1u)

Now, just add up (t) and (t') and differentiate the sum, we have

dT / dx = - 10 ((1/720 + x) - (1 / 10u-x))

720 + x = 10u - x

x = 5u - 360

Now, when u = 105 ft/sec then x = 165 ft and time = 3.43 seconds

when u = 115 ft/sec then x = 215 ft and time = 3.24 seconds

as time for u = 115 ft/sec is less than for u = 105 ft/s, the manager should encourage a relayed throw.