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An arrow is launched with an initial velocity of 7.6 m/s at an angle of 42 degrees above the horizontal. What is the horizontal component of the velocity at the highest point of the trajectory?

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Trancescient

Explanation:

At any point on the arrow's trajectory, the horizontal component of the velocity is the same. Therefore, the horizontal component of the velocity at the top of its trajectory is

[tex]v_x = v_{0x} = v_0\cos{42°} = (7.6\:\text{m/s})\cos{42°}[/tex]

[tex]\:\:\:\:\:=5.6\:\text{m/s}[/tex]

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