5. In 1947 Bob Feller, a pitcher for the Cleveland Indians, threw a baseball across the
plate at 98.6 mph or 44.1 m/s. For many years this was the fastest pitch ever measured.
If Bob had thrown the pitch straight up, how high would it have gone?

When the ball is thrown straight up, it follows a free fall motion, which is a uniformly accelerated motion with constant acceleration ([tex]g=9.8 m/s^2[/tex] towards the ground). Therefore, we can use the following suvat equation:

[tex]v^2-u^2=2as[/tex]

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

In this problem, we have:

u = 44.1 m/s is the initial vertical velocity of the ball

v = 0 is the final velocity when the ball reaches the maximum height

s is the maximum height

[tex]a=-g=-9.8 m/s^2[/tex] is the acceleration of gravity (downward, so negative)

Solving for s, we find the maximum height reached by the ball:

[tex]s=-\frac{u^2}{2a}=-\frac{44.1^2}{2(-9.8)}=99.2 m[/tex]

Learn more about free fall:

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5. In 1947 Bob Feller, a pitcher for the Cleveland Indians, threw a baseball across the plate at 98.6 mph or 44.1 m/s. For many years this was the fastest pitch ever measured. If Bob had thrown the pitch straight up, how high would it have gone?

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5. In 1947 Bob Feller, a pitcher for the Cleveland Indians, threw a baseball across the
plate at 98.6 mph or 44.1 m/s. For many years this was the fastest pitch ever measured.
If Bob had thrown the pitch straight up, how high would it have gone?