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 ski jumper starts from rest 55.0 m above the ground on africtionless track The skier flies of the track at an angle of =45.0o above the horizontal The height of the skier above the ground is H = 13.0 m (a)The height of the skier above the ground is H = (u2 * sin2/g)--------------(1) Here,g = 9.8 m/s2 From equation (1),we get u2= (2gH/sin2) or u = (2gH/sin2)1/2 =((2gH)1/2/(sin)) Substituting the values in the above equation,we get u = ((2 * 9.8 *13.0)1/2/(sin(45.0o)) or u = 22.57 m/s (b)Let maximum altitutde she attains after leaving the trackbe S,therefore,we get S = ut + (1/2)gt2 Here,t = (u * sin/g) or S = u * (u * sin/g) + (1/2) * g * (u *sin/g)2 or S = (3/2) * (u2 *sin2/2g) Substituting the values in the above equation,we get S = (3/2) * ((22.57)2 *sin2(45.0o)/(2 * 9.8)) or S = 19.5 m (c)The point where she lands relative to the end of the trackis R = (u2 * sin2/g) or R = ((22.57)2 * sin(2 *45.0o)/(9.8)) or R = 51.98 m or S = (3/2) * (u2 *sin2/2g) Substituting the values in the above equation,we get S = (3/2) * ((22.57)2 *sin2(45.0o)/(2 * 9.8)) or S = 19.5 m (c)The point where she lands relative to the end of the trackis R = (u2 * sin2/g) or R = ((22.57)2 * sin(2 *45.0o)/(9.8)) or R = 51.98 m

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