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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.

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

Added Thu, 25 Jun '15

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