As illustrated in Figure 12.14, consider a ski jumper moving down a track to acq

Question

As illustrated in Figure 12.14, consider a ski jumper moving down a track to acquire sufficient speed to accomplish the ski jumping task. The length of the track is C=25m and the track makes an angle theta =45degree with the horizontal. If the skier starts at the top of the track with zero initial speed, determine the takeoff speed of the skier at the bottom of the track using the work-energy theorem the conservation of energy principle the equation of motion along with the kinematic relationships. Assume that the effects of friction and air resistance are negligible.

Explanation / Answer

a) using work energy therorem ,

work done(work done by gravity ) = total energy (increase in K.E.)

mgh = mv^2 /2

9.8 x 25sin45 = v^2 /2

v = 18.61 m/s

b) conservation of energy

energy at top (P.E. + K.e.) = energy at bottom (P.E. + K.e)

mgh + 0 = 0 + mv^2 /2

9.8 x 25sin45 = v^2 /2

v = 18.61 m/s

c) a = gsin45 = 9.8sin45 = 6.93 m/s2

v^2 - u^2 = 2as

v^2 - 0 = 2 x 6.93 x 25

v = 18.61 m/s

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