Question

During a baseball game, a batter hits a high

Answer 1

Answer:

50.0 m

Explanation:

First of all, we can find the initial velocity of the ball, using the equation

v = u + at

where

v = 0 is the velocity of the ball at the highest position

u is the initial velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity

$t=\frac{6.37}{2}=3.19 s$ is the time the ball took to reach the maximum height (half of the time it remained in the air)

Solving for u,

$u=v-at=0-(-9.8)(3.19)=31.3 m/s$

Now we can find the maximum height using the other SUVAT equation:

$v^2-u^2 = 2ad$

where d is the maximum height. Solving for d,

$d=\frac{v^2-u^2}{2a}=\frac{0^2-(31.3)^2}{2(-9.8)}=50.0 m$