Aball is launched with an inital velocity of 4.47 m/s at an angle of 66° above the horizontal. calculate how long did it take the ball to return to its launching hike. use -9.80 m/s^2 for the acceleration due to gravity.

0.833 seconds

Explanation:

Given:

y₀ = 0 m

y = 0 m

v₀ = 4.47 sin 66° m/s

a = -9.80 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

0 = 0 + (4.47 sin 66°) t + ½ (-9.80) t²

0 = 4.08 t − 4.90 t²

0 = t (4.08 − 4.90 t)

t = 0, 0.833

It takes 0.833 seconds to return to its original launching height.

that would be b i just took the test : )

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