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A baseball is thrown with an initial velocity of 45.4 m/s at an angle of 31.2 ∘ .

Find the maximum height that the ball reaches using energy considerations. h = ___ m

Sagot :

Answer:

V (initial vertical velocity) = 45.4 sin 31.2 = 23.52 m/s

1/2 m V^2 = m g h      conservation of energy

h = V^2 / (2 g) = 23.52^2 / 19.6 = 28.2 m       max height

Check:

t = 28.2 / 9.8 = 2.88 sec    time to reach max height

h = 23.52 * 2.88 - 1/2 g 2.88^2 = 27.1 m

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