Modeling Baseball Strategy Question Details An outfielder picks up a ground ball

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Modeling Baseball Strategy
Question Details
An outfielder picks up a ground ball and wants to make a quick 150 ft throw to first base. To save time, he does not throw the ball directly to the first baseman. He skips the ball off the artificial surface at 75 ft so that the ball reaches first base after a bounce as shown in the figure on the next page. Let's examine this strategy. The equations

and
give the coordinates of the ball at time t seconds for an initial angle of ? degrees, initial velocity vo ft/sec, and initial height ho ft. Assume that the ball is thrown at 130 ft/sec from a height of 5 ft and the ball is caught at a height of 5 ft.

a) At what angle from horizontal must the ball be thrown to reach first base without a skip?
b) How long does it take the ball to get to first base without a skip?
c) At what angle from horizontal must the ball be thrown so that it strikes the ground at 75 ft?
d) How long does it take for the ball to reach the skip point at 75 ft?
e) Assume that the path of the ball is symmetric with respect to the skip point and double your answer to part (d) to get the time that it takes the ball to reach first base.
f) How much time is saved by skipping the ball?
g) How long would it take for the ball to reach first base if it could be thrown in a straight line?

h) Discuss the assumptions made for this model. Did we ignore anything that might affect our results?

Explanation / Answer

a.) the up and down coordinate of the location of the ball is
y = -16t^2 + vt sin theta. We want the first baseman to get the ball at the same height from which it was thrown, so we substitue in zero for y and get
16t^2 = vt sin theta
divide both sides by t
16t = v sin theta
t = v sin theta/16
Let's substitute this in x = vt cos theta
x = v^2 (1/16) sin theta cos theta
sin 2a = 2 cos(a)sin(a)
x = 130^2(1/16)(1/2)sin (2 theta)
theta = (1/2) ARCSIN(150*16*2/130^2) = 0.143994004 radians
theta = 0.143994004*180/PI degrees = 8.25 degrees
a.) theta = 8.25 degrees
Let's check this, it will take a little more than a second to go 150 ft. A ball falls 16 feet in a second, so we want 8 degrees (with no gravity) to put the ball a little more than 16 feet up
150feet sin8 = 21feet, so we are happy with our answer

b.) The nice thing about the horizontal speed is that it doesn't change.
distance = rate times time
t = d/r = 150ft/[130ft/sec cos 8.25 = 1.165912457s (I am keeping all the digits since it doesn't seem like we will save much time by using this technique)

c.) now we want to do mostly the same thing, but for 75 feet and for y = -5, since the ball will hit the ground. This makes the equation harder since y doesn't drop out.
If he throws the ball horizontally, it would fall for 75ft/(130ft/s) seconds. this would put it -5.32 below its starting point at 75 feet. This is so close (off by .32 feet) that I will try a few angles to see if I can get close enough.
Here I have the angle in the left column, cos angle next, time next, vtsin theta next, and finally, how far it fell after 75 feet
0.248 1 0.576928481 0.004328403 0.324633268 -5.000910293

0.249 1 0.576928525 0.004345856 0.32594229 -4.999602078
If the ball is thrown at 0.249 degrees up from the horizontal, it strikes the ground at 75 feet.
c.) 0.249 degrees
d.) it will take 0.5769 seconds
e.) 2 times 0.5769 = 1.15385705 seconds with the assumptions
f.) 1.165912457s - 1.15385705s = 0.012055407 s 0r 0.01s
g.) 150ft/130ft/s = 1.153846154
h.) We assumed an elastic collision with the field. We ignored wind resistance.
If the collision with the field is not elastic, the ball will not have enough height to get to the first baseman at 5' and it will lose some velocity. Wind resistance usually favors the more horizontal throw in term of range, so this might have less of an effect on the skipping trick than on the typical throw.

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