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

The maximum height Hmax is given by, Hmax=u2sin2/2g 13=u2sin245/2*9.8 u2=509.6 u=22.574m/s This is the initial speed which she leaves the track. The maximum altitude is H=Hmax+55=55+13=68m The vertical velocity the skier reaches the ground is v2-u2=2gh v2-0=2*9.8*68 v=36.5m/s From the maximum height the time taken to reach the groundis v-u=gt 36.5-0=gt t=36.5/9.8 t=3.725s The horizontal distance travelled is x=vx*t=22.574cos45*3.725 x=84.094m The total horizontal distance travelled is the half of range+84.094 Range R=u2sin2/g R=22.5742sin2*45/g R=51.998m The toatal distance travelled is R1=R/2+84.094 R1=25.99+84.094 R1=110.9325m Hence we get by it.

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