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A ball is thrown with velocity 30.m⋅s−1at an angle of 40.º below the horizontal. What is the magnitude of the horizontal component of the velocity? Give your answer to the nearest m⋅s−1 and do not include units with your answer.

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abidemiokin

Answer:

Explanation:

The horizontal component of the velocity is expresssed as;

Vx = Vcos θ

V is the velocity of the ball

θ is the angle of inclination to the horizontal

Given

V = 30m/s

θ = 40°

Vx = 30cos40°

Vx = 22.98

Hence the magnitude of the horizontal component of the velocity to the nearest m/s is 23

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