During a baseball game, a batter hits a high pop-up.

If the ball remains in the air for 4.59 s, how high does it rise? The acceleration of gravity is 9.8 m/s2 .

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 during the 4.59 seconds it remains in the air, we can use the equations of motion in vertical motion.

First, we need to determine the initial vertical velocity of the ball when it was hit. Assuming the ball is hit directly upwards, we can use the equation:

vf = vi + at

Since the ball reaches its maximum height, its final vertical velocity (vf) will be zero (as it momentarily comes to rest at the highest point). The acceleration (a) is the acceleration due to gravity (-9.8 m/s^2), and the initial vertical velocity (vi) is what we are trying to find.

0 = vi + (-9.8 * 4.59)

Rearranging the equation, we can solve for vi:

vi = 9.8 * 4.59

vi = 44.98 m/s (rounded to two decimal places)

Now, we can use another equation to calculate the height the ball rises during this time:

h = vi * t + (1/2) * a * t^2

where h is the height, vi is the initial vertical velocity, t is the time, and a is the acceleration due to gravity.

h = (44.98 * 4.59) + (1/2) * (-9.8) * (4.59^2)

h = 206.29 - 106.52

h = 99.77 meters (rounded to two decimal places)

Therefore, the ball rises approximately 99.77 meters during the 4.59 seconds it remains in the air.