Question

A ski jumper starts from rest 56.0 m above the ground on a frictionless track and flies off the track at an angle of 45.0° above the horizontal and at a height of 19.0 m above the ground. Neglect air resistance.

(a) What is her speed when she leaves the track?

(b) What is the maximum altitude she attains after leaving the track?

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

Answer 1

a) we know the formula for kinetic energy

K.E=1/2mv²....1)

and potential energy

P.E=mgh.........2)

so equating both 1 and 2 we get

1/2mv²=mgh

1/2v²=mg

v=2gh

h=56-19=37m

so

v=2*9.8*37=26.93m/s

b) we have the formula

v²=2gh

here v= initial velocity=26.93*sin 45=19.04m/s

(19.04m/s)²=2*9.8m/s²*h

h=18.5m

so the altitude max=18.5m+19m=37.5m

c) here the horizontal velocity is

v=26.93m/s*cos45=19.04m/s

we know the formula

h=1/2gt²

t²=2h/g

for upward time h=18.5m

t²=2*18.5m/9.8=3.7755s

t=1.9431s

for downward time h=37.5

t²=2*37.5/9.8=7.6531s

t=2.7664s

total time=1.9431+2.7664=4.7095s

we know distance =speed* time= 19.04*4.7095s=89.67m

answer a) 27m/s or 26.93m/s or 26.9 m/s b) 37.5 or 38m c) 90m or 89.67m or 89.7m

answered by: anonymous