Question #307600

Barney walks at a velocity of 1.7 meters/second on an inclined plane, which has an angle of 18.5° with the ground. What is the horizontal component of Barney’s velocity?


1
Expert's answer
2022-03-08T08:56:05-0500
vx=vcosθ=1.7m/scos18.5=1.6m/sv_x=v\cos\theta\\ =\rm 1.7\: m/s*cos18.5^\circ=1.6\: m/s


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