A baseball is thrown vertically upward (assumed to be the positive direction) with a speed of 19.3 m/s from a height of 1.18 m.

How high does the ball rise from its original position?

It rises until the initial kinetic energy is converted to gravitational potential energy.

The original height does not matter.

To find the maximum height reached by the ball, we can use the equations of motion.

First, let's identify the known quantities:
Initial velocity (u) = 19.3 m/s (upward)
Initial position (s) = 1.18 m
Final velocity (v) = 0 m/s (at the maximum height)
Acceleration (a) = -9.8 m/s^2 (assuming downward acceleration due to gravity)

We'll use the kinematic equation that relates displacement, initial velocity, final velocity, and acceleration:

v^2 = u^2 + 2as

Rearranging the equation to solve for displacement (s), we get:

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

Plugging in the values, we have:

s = (0^2 - 19.3^2) / (2 * -9.8)

Simplifying the equation gives:

s = (-373.49) / (-19.6)

s = 19.05 meters

Therefore, the ball rises approximately 19.05 meters from its original position.