A 0.5 kg ball is thrown upward at 12 m/s. How high does it go?

Step 1 — Identify the condition at the top

At the highest point the ball is momentarily at rest, so the final velocity is v = 0. That single fact is what makes the problem solvable.

Step 2 — Choose a method

Energy conservation is the shortest route: all the kinetic energy at launch becomes gravitational potential energy at the top.

½mv² = mgh → h = v² / (2g)

The mass cancels, which is why the 0.5 kg is not needed.

Step 3 — Substitute

h = 12² / (2 × 9.8) = 144 / 19.6 = 7.35 m

Cross-check with kinematics

Using v² = u² + 2as with v = 0, u = 12 and a = −9.8:

0 = 144 − 19.6h → h = 7.35 m

Both methods agree, which is the usual way to confirm that the sign of g was handled correctly.