Question

During a baseball game, a batter hits a high

pop-up.
If the ball remains in the air for 6.37 s, how
high does it rise? The acceleration of gravity
is 9.8 m/s2.

Answer 1

50.0 m

Step-by-step 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=(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=(v^2-u^2)/(2a)=(0^2-(31.3)^2)/(2(-9.8))=50.0 m`