How to play Straight Up
- Choose a throw and read the launch velocity v₀. Up is positive.
- Use v = v₀ − g·t and y = v₀·t − ½·g·t². At the top, v = 0.
- Type your answer, with a sign where the question asks for one, and press Play.
- Watch the arrows: the velocity shrinks, flips, and grows, while the acceleration stays −g. Compare your answer with the result.
Up and down with one acceleration
A ball thrown straight up has the same acceleration the whole time: a = −g = −9.8 m/s², downward. On the way up it slows by 9.8 m/s each second, stops for an instant at the top after t = v₀ ÷ g, and then speeds up downward. Its maximum height is y_max = v₀² ÷ (2g). The flight is symmetric: the way down takes as long as the way up, and the ball returns to the hand with velocity −v₀. At the top the velocity is zero, but the acceleration is not.
A focused learning model
The ball is a point thrown straight up from the hand, with g = 9.8 m/s² and no air resistance, so the motion is symmetric. Heights are measured from the hand. The top and the catch are solved exactly at t = v₀ ÷ g and t = 2v₀ ÷ g, and the velocity just before the catch is kept. The view and the graph use the same scales for every throw, so they never give an answer away.