Abaseball is launched straight up (vertically) next to a tall building with an initial velocity of 30.0 m/s[up]. the ball continues upward to a stop, and then falls until caught by a player leaning out of a window 14.0 m above the ground. how long is the ball in the air before the player catches it?

5.61 seconds

Explanation:

Given:

y₀ = 0 m

y = 14.0 m

v₀ = 30.0 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

14 = 0 + 30t − 4.9t²

4.9t² − 30t + 14 = 0

Solve with quadratic formula:

t = [ -b ± √(b² − 4ac) ] / 2a

t = [ 30 ± √(900 − 274.4) ] / 9.8

t = 0.509, 5.61

t = 0.509 is the time when the ball first reaches the height of 14.0 meters on the way up.  t = 5.61 is the time when the ball reaches the height of 14.0 meters on the way down.

Since we know the ball reaches its maximum height and begins falling again, we want the second time.  t = 5.61 seconds.

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0.19 m/s and 0.45 m/s