A baseball pitcher throws a ball at 90.0 mi/h in the horizontal direction. How far does the ball fall vertically by the time it reaches home plate, which is a horizontal distance of 60.5 ft away? (Express your answer in units of feet. Neglect air resistance.)

Question

Jakobe Davidson

Physics

1 March, 21:55

A baseball pitcher throws a ball at 90.0 mi/h in the horizontal direction. How far does the ball fall vertically by the time it reaches home plate, which is a horizontal distance of 60.5 ft away? (Express your answer in units of feet. Neglect air resistance.)

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Answers (1)

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Lucian · Answer 1

Vertical distance = 3.3803ft

Explanation:

First with the speed of the ball and the distance traveled horizontally we can determine the flight time to reach the plate:

Velocity = (90 mi/h) * (1 mile/5280ft) = 475200ft/h

Distance = Velocity * time⇒ time = 60.5ft / (475200ft/h) = 0.00012731h

time = 0.00012731h * (3600s/h) = 0.458316s

With this time we can determine the distance traveled vertically taking into account that its initial vertical velocity is zero and its acceleration is that of gravity, 9.81m/s²:

Vertical distance = (1/2) * 9.81 (m/s²) * (0.458316s) ²=1.0303m

Vertical distance = 1.0303m * (1ft/0.3048m) = 3.3803ft

This is the vertical distance traveled by the ball from the time it is thrown by the pitcher until it reaches the plate, regardless of air resistance.