a daring ski jumper landed with a velocity of 100m/s, 60 degrees below horizontal to the west. when she landed , she was 550 m horizontally west of the end of the ramp.the horizontal deceleration due to air friction was 1 m/s^2.what is the height of the ramp above the ground? with what speed and at what angle did she leave the ramp?
2007-04-28 15:21:38 · 1 answers · asked by rohandon c 1 in Science & Mathematics ➔ Physics
horizontal
v0^2 = 2500 - 2*550
v0^2 = 3600
v0 = 60 m/s
v = v0 + at
t = (50 - 60)/-1
t = 10 s
vertical
v = v0 + at
v0 = - 50√3 + 10*9.80662 (ignoring air resistance)
v0 = 11.46366 m/s
V = 61.08531 m/s @ 10.81663°
height of ramp above landing point:
s = s0 + v0t + (1/2)at^2
0 = s0 + 10*11.46366 - (1/2)(100)(9.80662)
s0 = 1,846.667 m
Assuming a 60° slope,
s0 = 894.0391 m at take-off point
2007-04-28 19:36:24 · answer #1 · answered by Helmut 7 · 0⤊ 0⤋