A ski jumper leaves the ski track moving in the horizontal direction with a speed of 25.0 m/ s. The landing incline below her falls off with a slope of 35.0 degrees.
Where does she land on the incline?
Estimate the time taken for the jump before landing
She lands on the incline after time (t) seconds at a point (x,y) which is horizontally "x" metres from the point of take off and vertically "y" metres below the point of take off.
x = (vx)t (where vx is the horizontal velocity)
x = 25t ......... equation 1
y = ut + (1/2)gt^2
y = 1/2 x 9.8 x t^2 (since u = 0 m/s as you correctly pointed out in your solution)
y = 4.9t^2 ......... equation 2
Now we solve equations 1 and 2 simultaneously
Let's square eq 1
x^2 = 625t^2
and then divide the two equations to get rid of t:
x^2/y = 625/4.9 ....... equation 3
Now we also know that tan? = y/x (where ? = angle of slope)
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