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Stella [2.4K]
3 years ago
14

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?
Physics
1 answer:
AleksAgata [21]3 years ago
5 0

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

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

\:\:\:\:\:=5.6\:\text{m/s}

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