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

(a) How high does it go?

At the very top of its flight, the ball has an instantaneous velocity of zero. We can plug v = 0 into the equation:

v² = v_o² + 2 a (x - x_o)

This gives:

0 = 144 + 2 (-9.8) x

Solving for x gives x = 7.35 m, so the ball goes 7.35 m high.

(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: The down half of the trip is a mirror image of the up half. While going up, if the ball passes through a particular height at a particular velocity (2 m/s up, for example), on its way down it will pass through that height at the same speed, with its velocity directed down rather than up. The up half of the trip takes the same time as the down half so we could just figure out how long it takes to reach its maximum height, and then double that to get the total time.

Another way to do it is simply to plug x = 0 into the equation:

x - x_o = v_o t + ½ a t²

This gives 0 = 12 t - 4.9 t²

A factor of t can be canceled out of both terms, leaving:

0 = 12 - 4.9 t, which gives a time of t = 12 / 4.9 = 2.45 s.

(c) How fast is it traveling when you catch it?

The answer has to be 12 m/s down, because of the mirror-image relationship between the up half of the flight and the down half. We could also figure it out using the equation:

v = v_o + a t

v = 12 - 9.8 (2.45) = -12 m/s.