A ball player catches a ball 3.24 s after throwing it vertically upward. With what speed did he throw it?

Doesn't seem like we know much here, but we can answer it. Let's talk about what we know. We know it takes 3.24 s for the ball to go up and drop back down again. We know that gravity is the only force acting after the ball leaves the hand, so a = 9.8 m/s^2 (we'll say it's negative in our equations because down being negative is intuitive). We also know that it stops moving for a brief moment at the top of the arc, where v = 0 m/s. Because gravity is the only force, and it slows it down on the way up at the same rate it speeds it up on the way down and the distance covered in upward and downward motion is the same, we can confidently say that it will reach the top of its arc (where v = 0 and it turns around) in half the total time it is in the air, so it takes 1.62 s to reach the peak. Now we can use a kinematics equation, let's use vf = vi + a*t, where vf is final velocity and is 0, vi is initial velocity and is some unknown v we need to solve for, a is acceleration and is - 9.8 m/s^2 and t is time and since this is just to the top of the arc, we'll use half the time so 1.62 s. We can solve for vi and plug stuff in like so: v = - a*t = - (-9.8m/s^2) * (1.62s) = 15.876 m/s.