During a baseball game, a batter hits a pop- up to a fielder 72 m away. The acceleration of gravity is 9.8 m/s^2. If the ball remains in the air for 5 s, how high does it rise?

TO ANYONE WHO SEES THIS - NEVERMIND, I GOT IT.

To find the height to which the ball rises, we can use the equation of motion for vertical motion:

h = ut + 0.5at^2

where:
h is the height,
u is the initial velocity (in this case, the velocity at which the ball leaves the bat),
t is the time, and
a is the acceleration due to gravity.

In this scenario, the ball is hit straight up, so its initial velocity is positive. We will assume that the initial velocity is known to be zero, as the question does not provide that information. Therefore, the equation simplifies to:

h = 0.5at^2

Substituting the given values:
a = 9.8 m/s^2 (acceleration due to gravity)
t = 5 s (time the ball remains in the air)

h = 0.5 * 9.8 * (5)^2
h = 0.5 * 9.8 * 25
h = 122.5 m

Therefore, the ball rises to a height of 122.5 meters.