Question

1. -/1 points one of the fastest - recorded pitches in major - league baseball, thrown by tim lincecum in 2009, was clocked at 101.0 mi/h. if a pitch were thrown horizontally with this velocity, how far would the ball fall vertically by the time it reached home plate, 60.5 ft away? details my notes ask your teacher

Explanation:

Step1: Convert velocity to ft/s

1 mile = 5280 ft, 1 hour = 3600 s. So, $101.0$ mi/h is $101.0\times\frac{5280}{3600}\text{ ft/s}\approx148.13\text{ ft/s}$.

Step2: Find time to reach home - plate

The distance to home - plate is $d = 60.5$ ft. Using the formula $t=\frac{d}{v}$, where $v$ is the horizontal velocity. So, $t=\frac{60.5}{148.13}\text{ s}\approx0.408\text{ s}$.

Step3: Find vertical fall

The vertical motion is a free - fall with initial vertical velocity $v_{0y}=0$ and acceleration $a = g=32.2\text{ ft/s}^2$. Using the formula $y = v_{0y}t+\frac{1}{2}at^{2}$, since $v_{0y} = 0$, we have $y=\frac{1}{2}\times32.2\times(0.408)^{2}\text{ ft}$.
$y=\frac{1}{2}\times32.2\times0.166464\text{ ft}\approx2.69\text{ ft}$.

Answer:

$2.69$ ft