Colby, of mass 74 kg, decides to go ziplining down a zipline that is 100 m long on the hypotenuse and declines at an angle of 10°. If he starts from rest at what speed will he land at the bottom if friction can be ignored?
sin10°=x/100→x=100⋅sin10°=17.4 (m)\sin10°=x/100\to x=100\cdot\sin10°=17.4\ (m)sin10°=x/100→x=100⋅sin10°=17.4 (m)
mgx=mv2/2→v=2gx=2⋅9.8⋅17.4≈18.5 (m/s)mgx=mv^2/2\to v=\sqrt{2gx}=\sqrt{2\cdot9.8\cdot17.4}\approx18.5\ (m/s)mgx=mv2/2→v=2gx=2⋅9.8⋅17.4≈18.5 (m/s) . Answer
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