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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

Obtain the velocity when the jumper lands.
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Added Thu, 25 Jun '15
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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)
y = x tan35

Now substitute this into equation 3 above to get
x = (625/4.9) x tan35
x = 127.55 x tan35
x = 89.31 m
and
y = 89.31 x tan35
y = 62.54 m
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Added Thu, 25 Jun '15
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