Question #252021

A sky jumper speeds down a slope and exits the ski track in a horizontal direction with a velocity of 25 m/s where the landing incline falls off with a slope of 33°. Solve the following:

A. What is the half time of flight of the sky jumper?

B. What is the range of the sky jumper when it lands?


Expert's answer

h=gt2/2h=gt^2/2


l=vtl=vt


h=l⋅tan⁡33°→gt2/2=vt⋅tan⁡33°h=l\cdot\tan33°\to gt^2/2=vt\cdot\tan33°


t=2vtan⁡33°/g=2⋅25⋅tan⁡33°/9.8=3.31 (s)t=2v\tan33°/g=2\cdot25\cdot\tan33°/9.8=3.31\ (s)


t1/2=1.66 (s)t_{1/2}=1.66\ (s) . Answer


l=25⋅3.31=82.75 (m)l=25\cdot3.31=82.75 \ (m) . Answer

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