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An arrow is launched from a bow with an initial horizontal velocity of 40

m/s and an initial vertical velocity of 62 m/s. What is the total initial vector

velocity of the arrow?


Sagot :

Answer:

V = (Vx^2 + Vy^2)^1/2 = (40^2 + 62^2)^1/2

V = 73.8 m/s

tan theta = Vy / Vx = 62/40 = 1.55

theta = 57.2 deg

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