MATH SOLVE

7 months ago

Q:
# A major league baseball pitcher throws a pitch that follows these parametric equations:x(t) = 143ty(t) = β16t2 + 5t + 5.Recall that the speed of the baseball at time t is s(t)=β [x '(t)]2 + [y ' (t)]2 ft/sec. What is the speed of the baseball (in mph) when it passes over homeplate?

Accepted Solution

A:

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