A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 17.0 m/s when the hand is 1.50 m above the ground. How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)

The ball is 3.16 seconds in the air before it hits the ground.

Explanation:

Upward:

v = v o - g t

0 m/s = 15 m/s - 9.81 m/s² · t

15 m/s = 9.81 m/s² · t

t = 15 m/s : 9.81 m/s²

t 1 = 1.53 s

h = v o · t - g t² / 2  + 2 m

h = 22.95 m - (9.81 m/s² · 2.3409 s²) / 2 + 2 m

h = 11.475 m + 2 m = 13.475 m

Downward:

13.475 m = 9.81 · t² / 2

t = √(13.475 · 2 ) / 9.81

t 2 = 1.66 s

t = t 1 + t 2 = 1.53 s + 1.66 s = 3.19 s

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