Question

5. In 1947 Bob Feller, a pitcher for the Cleveland Indians, threw a baseball across the

Answer 1

The maximum height reached by the ball is 99.2 m

Explanation:

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

$v^2-u^2=2as$

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

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

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

$s=-\frac{u^2}{2a}=-\frac{44.1^2}{2(-9.8)}=99.2 m$

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