Question

one of the fastest recorded pitches in major league baseball. thrown by billy wagner in 2003, was clocked at 101.0 mi/h. if a pitch were thrown horizontally with this velocity, how far would the ball fall vetically by the time it reached home plate, 60.5 ft away. Report answer in units of ft.

Answer 1

To solve this problem we will first standardize the units given. We will apply the cinematic equations of motion for which with the given information we will find through the displacement in X, the time of the ball in fulfilling its course. From time on, we will look for vertical displacement Y.

$101 \frac{mi}{h} (\frac{1609.344m}{1mi})(\frac{1h}{3600s}) = 45.15m/s$

$60.5ft (\frac{0.3m}{1ft}) = 18.1m$

We'll use the kinematic equations:

$\Delta x = vt$

$18.1 = (45.15104)t$

$t = 0.40s$

Now the vertical displacement would be,

$\Delta y = v_0 t +\frac{1}{2} at^2$

Here,

$v_0$ = Initial velocity (Zero at our case)

a = Acceleration (Gravity at our case)

t = time

Replacing we have,

$\Delta y = v_0 t +\frac{1}{2} at^2$

$\Delta y = (0)(0.4) +\frac{1}{2} (9.8)(0.4)^2$

$\Delta y = 0.784m$

Converting to feet we have,

$0.784m (\frac{1ft}{0.3m})  = 2.61ft$

Therefore the ball fall vertically around to 2.61ft