Answer to Question #81485 in Mechanics | Relativity for Yengo

Question

A ski-jumper leaves the ski track moving in horizontal direction witha speed of 25m/s. The landing incline below him falls of with a slope of 35.0 degrees. Where does he land on the incline?

Answer 1

Loss of height by skier:
h_j=(gt^2)/2
The height of the slope at the point x:
h_j=x*tg α
Coordinate x:
x=Vt
Coordinate y:
〖y=h〗_j=h_t
We obtain and transform equation:
(gt^2)/2= x*tg α
(gt^2)/2= Vt*tg α
gt/2= V*tg α
We allocate time of flight of the skier:
t=2V/g*tg α=2*25/9.81*0.7=3.57s
And we get the coordinates of the landing:
x=(2V^2)/g*tg α=(2*625)/9.81*0.7=89.2m
y=(gt^2)/2= x*tg α=89.2*0.7=62.44m
Flight distance:
s=√(x^2+y^2 )=108.9m