You throw a ball straight up. It leaves your hand at 12.0 m/s.
(a) How high does it go?
(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?
(c) How fast is it traveling when you catch it?
Origin = height at which it leaves your hand
Positive direction = up
At the very top of its flight, the ball has an instantaneous velocity of zero. We can plug v = 0 into the equation:
v2 = vo2 + 2 a (x - xo)
0 = 144 + 2 (-9.8) x
Solving for x gives x = 7.35 m, so the ball goes 7.35 m high.