A ski jumper starts from rest 44.0 m abovethe ground on a frictionless track and

Question

A ski jumper starts from rest 44.0 m abovethe ground on a frictionless track and flies off the track at anangle of 45.0° above the horizontal and at a height of11.0 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

a) h1=44m, h2=11m, so h=44-11=33m. Then we use conservationof energy to find that PEi=KEf or mgh=1/2mv2 orgh=1/2v2 and then v=(2gh)=25.43 m/s. b) Now we have Vo=25.43m/s and =45, so max height occurs whenvy=0 then we use Vy=VoSin-gt which becomes0=25.43Sin45-(9.8)t and then we find that t=1.835s. Now we use theequation y=VoSint-1/2gt2 and t=1.835s , and we also need to use the height of the ramp as an initialheight, so y-11=25.43Sin45(1.835)-4.9(1.835)2˜27.5meters above the ground. c) x=VoCost and thenx=25.43(Cos45)(1.835)=32.3meters from the ramp

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