(b) If, when the ball is on the way down, you catch it at the same height at which you let it go, how long was it in flight?

Method 1: While going up, if the ball passes through a particular height at a particular velocity, on its way down it will pass through that height at the same speed, with its velocity directed down rather than up. The down half of the trip is a mirror image of the up half. We could just figure out how long it takes to reach its maximum height, and then double that to get the total time.

Another method is to simply plug y = 0 into the equation:

y - y_o = v_o t + ½ a t²

This gives 0 = v_o t + ½ a t²

Cancelling a factor of t:

0 = v_o + ½ a t

t

=

-2vo a

=

-24 -9.8

=

2.45 s