A 140 g, 90.0 mph fastball comes across the plate very nearly horizontally and i

Question

A 140 g, 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 ball.
(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 Fmax.
(d)Find the average acceleration of the ball.
(e)For some extra 2-D motion practice, does this ball clear the outfield fence, 410 ft away?

Explanation / Answer

m = 140 g = 0.14 kg , ux = 90 mph = 144.841 kmph = 40.234 m/s , uy = 0
vx = -(90cos 30) mph = -(144.841cos 30) kmph = -(40.234cos 30) m/s = - 34.84 m/s

vy = 40.234 sin(30) m/s =20.117 m/s

impulse,I = change in momentum = m(v - u) = m( vx - ux ) i + m( vy - uy ) j = 0.14( (-34.84-40.234) i + (20.117-0) j ) = -10.51 i + 2.82 j

magnitude of impulse, I = [ (-10.51)2 + (2.82)2 ] = 10.88 kg.m/s = 10.88 N.s ans(a)

since, impulse,I=F*t, for t=0.0008 s, F=13602.19 N ans(b)

Impulse

Fmax = 13088094.22 N ans(c)

averge acceleration,aavg =|(v-u)|/t =|(I/m)|/t =|( -75.074 i + 20.117 j)|/0.0008=97153.223 m/s2 ans(d)

distance, s=-410 ft = -124.968 m

time taken to reach fence, t'=s/vx = 3.59 s

height of ball at this t' with acceleration due to gravity, g ,h=vyt' - 0.5gt2 = 20.117*3.59 - 4.9*3.592 =9.07m

yes the ball will clear outfield fence ans(e)

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