During a baseball game, a batter hits a pop-up to a fielder 92m away. The acceleration is 9.8 m/s^2. If the ball remains in the air for 5.8 seconds, how high does it rise?

Tf = Tr = 5.8/2 = 2.9 s.

Tr = Rise time.
Tf = Fall time.

h = 0.5g*Tf^2

To find the height that the ball rises, we can use the kinematic equation:

h = v0t + (1/2)at^2

Where:
h = height (what we want to find)
v0 = initial vertical velocity (0 m/s since the ball is hit straight up)
t = time (5.8 seconds)
a = acceleration (9.8 m/s^2, directed upwards)

First, let's find the initial vertical velocity (v0):

v0 = g × t

Where g is the acceleration due to gravity (9.8 m/s^2) and t is the time (5.8 seconds).

v0 = 9.8 m/s^2 × 5.8 seconds
v0 = 56.84 m/s

Now, we can substitute the values into the first equation to find the height (h):

h = (0 m/s × 5.8 seconds) + (1/2)(9.8 m/s^2)(5.8 seconds)^2
h = 0 + (1/2)(9.8 m/s^2)(33.64 seconds^2)
h = (1/2)(316.3 m)
h = 158.15 m

Therefore, the ball rises approximately 158.15 meters.