- #1

atxjoe512

- 3

- 0

I recently had a question on an exam that went something like this. "A ball is thrown directly up from a building 59.4 meters high. It barely misses the building on its way down and hits the ground 4 seconds after being thrown. What is the final velocity of the ball?"

I got the answer correct through a plug and chug method of using the formulas. However, I don't understand.

How does one simple formula account for both the ball moving up and then coming back down? Why don't you have to use two formulas... The first showing the final height after the ball was thrown from a height of 59.4m, and the next formula taking that maximum height to calculate the balls final velocity when it hits the ground.

I'm confused how the formula incorporates both the going up and going down of the ball.

Please help! I'd like a conceptual description if possible.