Question

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?

Answer 1

Answer: 0.48 s

Distance covered,
`d= 60 ft`

In SI units,

`1 ft = 0.3048 m`

`60 ft= 60 ft * 0.3048 m/ft=18.28 m`

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

`Time elapsed= (Distance)/(speed)`

`t=(18.28 m)/(38.16 m/s)=0.48 s`

Answer 2

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