2. A 140g. 90.0 mph fastball comes across the plate very nearly horizontally and
ID: 1320050 • Letter: 2
Question
2. A 140g. 90.0 mph fastball comes across the plate very nearly horizontally and is batted at 90.0 mph at a 30.0 degree angle (above the horizontal) toward center field. (a) Determine the impulse (in N s) acting on the hall. (b) If the impact time is 0.00080 s, determine the average force on the ball. (c) The force on the ball as a function of time can be modeled as F(t) =Fmax sin^2(4000t). Using this function, assuming it will give the same impulse as the average force, find F max. (d) Find the average acceleration of the ball. (e) For some extra 2-D motion practice, does this hall clear the 7 foot tall outfield knee, 410 ft away?Explanation / Answer
As the ball interacts with the bat, it changes its momentum vector and kinetic energy vector from 90 mph toward the batter to 90 mph away from the batter at a 30 degree angle above the horizontal.
Let's get this into consistent units:
1 mph = 1.609 km/h = 1609 m/h = 0.4469 m/s
90 mph = 40 m/s
a) The Impulse is equal to the total change in the object's momentum.
The change in velocity vector is A - B.
A = -40 x
B = 40 * cos(30) x + 40 * sin(30) y
= -74.64x + 20y
The magnitude is sqrt (74.64^2 + 20^2) = 77.27 m/s
The change in momentum is 140g * 77.27 m/s = 10.82 kgm/s (answer)
b) Force is the change in momentum per unit time, so the average force is
10.82kgm/s / 0.00080s = 13525 kgm/s^2 = 1.353e4 N (answer)
c) The total impulse is equal to the area under the curve of
Fmax*sin^2(4000t)
A useful trig identity is sin^2(x) + cos^2(x) = 1
The area under the curve of sin^2(x) + cos^2(x) from 0 to pi is equal to pi.
The area under the curve of sin^2(x) from 0 to pi is pi/2.
We're only interested in one period of the sin^2 curve, so we want the area under one loop of the curve. But the curve has been compressed by a factor of 4000. One whole cycle runs, instead of from x = 0 to x = pi, from x = 0 to x = pi/4000. The area under that loop of the curve is pi/8000.
So the totalized force is Fmax*pi/8000 = force times time = impulse.
Fmax * pi/8000 = 10.82
Fmax = 10.82*8000/pi = 2.755e4 N
d) F(ave) = ma(ave) = m (v2 - v1)/t
1.353e4 N = 0.14*a(ave)
a(ave) = 1.353e4N/0.14kg = 9.664e4 m/s^2 (answer)
e) The outfield fence is 410 feet away = 125m.
At an x component velocity of 37.27 m/s, it will take 3.354 seconds to reach the wall.
The y component of velocity is 20 m/s upward. g = 9.81 m/s^2, so v = at.
During that time, the ball will move up and down a distance given by s = s(0) + vt - 1/2*a*t^2
a = 9.81 m/s^2
s = s(0) + 20*3.354 - 1/2 * 9.81 * 3.354^2 = s(0) + 67.08 - 55.09 = s(0) + 11.99 m
If s(0) = 0 (meaning instead of a baseball bat, you're using a golf club), the ball will clear a 11.99 meter fence.
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