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In major league baseball, the pitcher's mound is 60 feet from the batter.

If a pitcher throws a 91 mph fastball, how much time elapses from when the ball leaves the pitcher's hand until the ball reaches the batter?

Verified Answer :

scme1702

Converting the speed from miles per hour to feet per second:
91 mph = 133.5 feet / second

Time = distance / speed
Time = 60 / 133.5
Time = 0.45 seconds

Answer: 0.48 s

Distance covered, \(d= 60 ft\)

In SI units,

\(1 ft = 0.3048 m\)

\(60 ft= 60 ft \times 0.3048 m/ft=18.28 m\)

Speed, \(v= 91 mph= 91 miles/hr \times 1.61 km/miles \times \frac{5}{18}\frac {m/s}{km/hr}=38.16 m/s}\)

\(Time elapsed= \frac{Distance}{speed}\)

\(t=\frac{18.28 m}{38.16 m/s}=0.48 s\)




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