An object is thrown vertically upward with a speed of 22.4 m/s. How high does it rise?

the speed at the maximum point is zero.

Vf^2=Vi^2+2gh g= -9.8m/s solve for h.

how would you rearrange it from

0 = (22.4m/s)^2+2(-9.8m/s)h

To find out how high the object rises, we need to use the equations of motion. In this case, since the object is thrown vertically upward, we can use the equation that relates the final velocity (v), initial velocity (u), acceleration (a), and displacement (s) in the vertical direction:

v^2 = u^2 + 2a * s

First, let's identify the given values:

Initial velocity (u) = 22.4 m/s (object is thrown upward)
Acceleration (a) = -9.8 m/s^2 (negative because the force of gravity is acting in the opposite direction to the motion)
Final velocity (v) = 0 m/s (at the highest point, the object momentarily stops)
Displacement (s) = ?

Now, we can substitute the known values into the equation:

0^2 = (22.4)^2 + 2*(-9.8) * s

Simplifying the equation:

0 = 501.76 - 19.6s

Rearranging the equation to solve for the displacement (s):

19.6s = 501.76
s = 501.76 / 19.6
s ≈ 25.62 meters

Therefore, the object rises approximately 25.62 meters before reaching its highest point.