Answer to Question #157961 in Mechanics | Relativity for John Bittinger

Question

The distance from the pitcher’s mound to home plate is 60 feet 6 inches (18.44

m). The fastest recorded pitch is 104.8 mph (46.85 m/s). If the pitch was thrown

with a release angle of 0o relative to the horizontal, what is the vertical

displacement of the ball between ball release on the mound and home plate?

Assume the ball is released 18.44 m from home plate.

Answer 1

Explanations & Calculations

• What is asked here is very basic as it is the vertical height is travels during its flight.
• Instead of throwing it above the horizontal, it is thrown horizontally at the beginning which moves the pitch in a trajectory completely below the horizontal line drawn at the mound.
• Gravity pulls the pitch down which leads the motion into a trajectory.
• So the vertical displacement is the height between the mound & the base plate.

• Applying motion equations both horizontally & vertically, the depth it it travels can be found.
• Horizontally,

$\qquad\qquad \begin{aligned} \small \rightarrow S&= \small ut+\frac{1}{2}at^2\\ \small 18.44 m&=\small 46.85ms^{-1}\times t+\frac{1}{2}\times9.8ms^{-2}\times t^2\\ \small 4.9t^2+46.85t-18.44&= \small 0\\ t&=\begin{cases} \small 0.3786s\\ \small -9.9398s \end{cases}\\ \small \therefore t&= \small 0.3786s \end{aligned}$

• Vertically,

$\qquad\qquad \begin{aligned} \small \downarrow S&= \small ut+\frac{1}{2}at^2\\ \small h&=\small 0+\frac{1}{2}\times (+9.8ms^{-2})\times (0.3786s)^2\\ &=\small \bold{0.702m}. \end{aligned}$

• That is the vertical height the pitch travels.