Question #139198
A stone is thrown with an initial velocity of 55 ft/
s, at an angle of 23° with the horizontal. Find the
vertical and horizontal components of the velocity
1
Expert's answer
2020-10-20T07:18:41-0400



Vx=Vcos(a)=55cos(230)50.63\vert{V_x}\vert=\vert{V*\cos(a)}\vert = 55*\cos(23^0)\approx50.63 ft/s

Vy=Vcos(90a)=Vsin(a)=\vert{V_y}\vert=\vert{V*\cos(90-a)}\vert = \vert{V*\sin(a)}\vert = 55sin(230)55*\sin(23^0)\approx 21.49 ft/s

Answer:

21.49 ft/s vertical component of the velocity

50.63 ft/s horizontal component of the velocity


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