A major league baseball pitcher throws a pitch that follows these parametric equations:

x (t) = 143t

y (t) = - 16t2 + 5t + 5.

Recall that the speed of the baseball at time t is

s (t) = √ [x ' (t) ]2 + [y ' (t) ]2 ft/sec.

What is the speed of the baseball (in mph) when it passes over homeplate?

97.67 mph The distance between the pitches plate and homeplate is 60.5 ft. So the time t at which the ball passes over home plate will be 60.5 = 143t 0.423 = t Now calculate the first derivative of each equation. So x (t) = 143t x' (t) = 143 y (t) = - 16t^2 + 5t + 5 y' (t) = - 32t + 5 So at 0.423 seconds, the respective velocities will be x' (0.423) = 143 y' (0.423) = - 32 * 0.423 + 5 = - 13.536 + 5 = - 8.536 And the speed will be sqrt (143^2 + (-8.536) ^2) = sqrt (20449 + 72.8633) = sqrt (20521.86) = 143.2545 ft/s Now convert from ft/s to mile/hour 143.2545 ft/s * 3600 s/hour / 5280 = 97.67 mile/hour = 97.67 mph