Question #251265

a sky jumper speeds down a slope and exits the ski track in horizontal direction at a velocity of 25 m per second where the landing inclined falls of the slope of 33 °

solve the following

what is the halftime of flight of the sky jumper

what is the range of the sky jumper when it lands


Expert's answer

Determine the height between the point of jump and point of landing:


h=gt22.h=\frac{gt^2}2.

Determine the range:


R=vt.R=vt.

The relation of R to h is


h=Rtan⁡θ=vttan⁡θ. vttan⁡θ=gt22, t=2vtan⁡θg=3.31 s,h=R\tan\theta=vt\tan\theta.\\\space\\ vt\tan\theta=\frac{gt^2}2,\\\space\\ t=\frac{2v\tan\theta}{g}=3.31\text{ s},

the halftime is 1.66 s, the range is


R=25⋅3.31=82.75 m.R=25·3.31=82.75\text{ m}.


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