Problem 2 10 pts From a height ht9 m the skier leaves the ra

Problem 2 (10 pts) From a height ht-9 m, the skier leaves the ramp at a speed of 17.966 m/s and an angle 0A-18 deg with the horizontal and strikes some distance down the slope at B. If the slope of the slope (rise/run)-4/5, determine strike distance, dist.

Solution

Let us assume that the skier travels a horizontal distance of x metres before hitting the ramp.

Further, the horizontal velocity of the skier will suffer no acceleration, hence will remain same as 17.966 cos18 = 17.0867 m/s

The time required for it to travel the horizontal distance of x = x / 17.0867 seconds

Now for vertical direction we have: y = 9 + 0.8x

9 + 0.8x = -17.966 sin18 + 0.5*9.81*x^2/17.0867^2

or, 9 + 0.8x =-0.325x +  0.0168x^2

or, 0.0168x^2 - 1.125x - 9 = 0

That is, x = [1.125 +/- sqrt(1.125*1.125 + 36*0.0168)] / 2*0.0168 = (1.125 +/- 1.368) / 2 = 1.2465 metres

Hence, y = 9+ 1.2465 = 10.2465 metres

Therefore the distance along the slope = sqrt (x^2 + y^2) = 10.322 metres