Last week we considered problems of motion with constant speed along a path, or constant velocity in up to two dimensions. This week, we go one step further, to consider motion with constant acceleration in one or two dimensions. A little bit of calculus sets the stage for the discussion. Armed with the general formulas above, we can solve any problem involving constant acceleration.
Problems can be simple or compound.
We are not concerned at all with horizontal motion in this problem. We aren't even told if John landed directly below the end of the diving board or not. The problem therefore is in just one dimension, the vertical. We want to find when the ball returns to my hand, and therefore when this expression vanishes. Notice that the time for the ball to go up and come back down is exactly twice the time to go up, or in other words, the time to go up is the same as the time to come back down.