What is the ball’s velocity when he catches it?
We can determine the answer to this question without any math at all. We know the initial velocity,
m/s, and the acceleration due to gravity,
m/s
2, and we know that the displacement is
x = 0 since the ball’s final position is back in the student’s hand where it started. We need to know the ball’s final velocity,
v, so we should look at the kinematic equation that leaves out time,
t:
Because both
x and
are zero, the equation comes out to
But don’t be hasty and give the answer as
12 m/s: remember that we devised our coordinate system in such a way that the down direction is negative, so the ball’s final velocity is
–12 m/s.
How high does the ball travel?
We know that at the top of the ball’s trajectory its velocity is zero. That means that we know that
= 12 m/s,
v = 0, and
m/s
2, and we need to solve for
x:
How long does it take the ball to reach its highest point?
Having solved for x at the highest point in the trajectory, we now know all four of the other variables related to this point, and can choose any one of the five equations to solve for t. Let’s choose the one that leaves out x:
Note that there are certain convenient points in the ball’s trajectory where we can extract a third variable that isn’t mentioned explicitly in the question: we know that x = 0 when the ball is at the level of the student’s hand, and we know that v = 0 at the top of the ball’s trajectory.
