A ski jumper starts from rest 55.0 mabove the ground on a frictionless track and

Question

A ski jumper starts from rest 55.0 mabove the ground on a frictionless track and flies off the track atan angle of 45.0° above the horizontal and at a height of13.0 m above the ground. Neglect airresistance. (a) What is her speed when she leaves thetrack?
1 m/s

(b) What is the maximum altitude she attains after leaving thetrack?
2 m

(c) Where does she land relative to the end of the track?
3 m (a) What is her speed when she leaves thetrack?
1 m/s

(b) What is the maximum altitude she attains after leaving thetrack?
2 m

(c) Where does she land relative to the end of the track?
3 m

Explanation / Answer

To answer part a use: mgH=1/2mv2+mgH' H=55 and H'=13 divide both sides by m to get rid of it since we don't know themass of the skier 9.8X55=.5v2+9.8X13 539=.5v2+127.4 v2=823.8 v=28.7m/s For part b: mgH'+1/2mv2=1/2m(vcos45)2+mgH'' again, divide by m 9.8(13)+1/2(28.72)=1/2(28.7cos45)2+9.8(H'') 127.4+411.845=205.9225+9.8H'' H''=34.0125m Sorry but I don't know how to answer part c!

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