During a baseball game, a batter hits a popup to a fielder 83 m away. The acceleration of gravity is 9.8 m/s 2 . If the ball remains in the air for 5.8 s, how high does it rise? Answer in units of m.

41.2 m.

Explanation:

It takes half the time that is (5.8/2) = 2.9 seconds, for the ball to reach its apex.

Given:

S = 83 m

t = 5.8 s

vf = 0 m/s

a = - g

= - 9.81 m/s^2

Equations of motion:

i. vf = vi + a * t

ii. h = vi * t + 1/2 (a * t²)

Using the i. equation of motion:

0 m/s = vi - (9.8 m/s²) (2.9 sec)

vi = 28.4 m/s.

Using the ii. equation of motion:

h = (28.4 * 2.9 ) - 1/2 (9.8 * (2.9)²)

= 82.4 - 41.2

= 41.2 m.

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