Video narration
A ball is dropped from rest from 18 meters above the ground. We ignore air resistance, so the acceleration is constant: g equals 10 meters per second squared downward. The question asks: at what height above the ground does the ball’s speed equal the magnitude of its acceleration, “in the same set of units”? Here that means we compare numbers: set the speed v equal to 10. Use the free-fall kinematics relation v squared equals 2 g times the distance fallen, since the initial speed is zero. So 10 squared equals 2 times 10 times s. That’s 100 equals 20 s, so s equals 5 meters fallen. The ball started at 18 meters, so its height above the ground then is 18 minus 5, which is 13 meters. So the ball’s speed is 10 meters per second at 13 meters above the ground.