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/s
2
.
Answer in units of m

Tf = T/2 = 6.37/2 = 3.19 s. = Fall time.

h = 0.5*g*Tf^2 = 4.9*3.19^2 =

vf-vi=at, so find that negative number then divide by two and it will give u |vi|

make it positive
vf =0 at the top, so
0^2=vi^2 + 2a ∆x
then
-(vi^2)/(2a){should be 19.6}= ∆x

To determine the height the ball rises, we can use the kinematic equation for vertical motion:

h = (1/2) * g * t^2

Where:
h is the height
g is the acceleration due to gravity (9.8 m/s^2)
t is the time the ball remains in the air (6.37 s)

Substituting the values into the equation:

h = (1/2) * 9.8 m/s^2 * (6.37 s)^2

Simplifying the equation:

h = 0.5 * 9.8 m/s^2 * 40.5369 s^2

h = 198.3854 m

Therefore, the ball rises to a height of 198.3854 meters.