Answer to Question #81228 in Physics for NEHA

Question

4. A ski jumper travel down a slope and leaves the ski track moving in the horizontal direction with a speed of 25m/s as in figure-2. The landing incline below him falls off with a slope of 35⁰. Where does he land on the incline?

Answer 1

She lands on the incline after time (t) seconds at a point (x,y) which is horizontally "x" meters from the point of takeoff and vertically "y" meters below the point of takeoff.
x=vt= 25t
y = (1/2)gt^2 =1/2 9.8 t^2=4.9t^2.
Thus,
x^2/y =625/4.
We know that
y = x tan35
Now substitute this into equation above to get
x = (625/4.9)tan35=89.3 m.
y = 89.3 tan35 = 62.5 m.