A ball is launched with an initial velocity of 4.47 m/s at an angle of 66° from the horizon. a. What is the maximum height the ball attained?
b. How long did it take the ball to return to the launching height?
c. How far away did it land?

See answers below

Explanation:

a.

(vf)² = (vi)² + 2gy

0 = (4.47sin66°)² + 2(-9.8)y

y = 0.00072 m

b.

vf = vi + gt

0 = (4.47sin66°) + (-9.8)t

t = 0.012 s

0.012 x 2 = 0.024 s

c.

x = (vi)t + (1/2)at²

x = (4.47cos66°)(0.024) + 0

x = 0.107 m

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