Question

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?

Answer 1

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

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

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