a person throws a ball upward into the air with an initial velocity of 20.0 m/s, calculate how high it goes

It rises to height H such that the potential energy gain equals initial kinetic energy.

M g H = (1/2) M V^2

H = V^2/(2 g)

To calculate how high the ball goes, we can use the principles of projectile motion. We need to determine the maximum height reached by the ball.

The key information given is the initial velocity of the ball when it was thrown upward, which is 20.0 m/s.

To solve this problem, we can use the following kinematic equation:

v^2 = u^2 + 2as

where:
v = final velocity (which is 0 m/s at the maximum height)
u = initial velocity
a = acceleration due to gravity (approximately -9.8 m/s^2)
s = displacement (the height we want to find)

Rearranging the equation, we get:

s = (v^2 - u^2) / (2a)

Substituting the given values into the equation:

s = (0^2 - 20.0^2) / (2 * (-9.8))

Now, we can calculate:

s = (0 - 400) / (-19.6)
s = -400 / -19.6
s = 20.41 meters

Therefore, the ball reaches a height of approximately 20.41 meters. Note that the negative sign indicates the direction of the displacement is upward.