Rearranging terms, the solutions for f and τ in terms of V , g and θ are as follows. f = 2 V 2 tan θ g cos θ and τ = 2 V tan θ g (b) The ski jumper’s angle relative to the horizontal is simply the arctangent of the ratio of the jumper’s vertical to horizontal velocity components, | ˙ z/ ˙ x | , wherefore φ =tan − 1 p g τ V Q − θ where we subtract the slope angle, θ ,sotha t φ is the jumper’s angle relative to the slope. From the solution to Part (a), we have g τ V = 2 V tan θ V =2tan θ Therefore, the ski jumper’s angle relative to the slope at the moment the jumper lands is φ =tan − 1 (2 tan θ ) − θ (c) In order to find the maximum value of φ , we differentiate the result of Part (b) with respect to θ , i.e., d φ d θ = 1 1+4tan 2 θ d d θ (2 tan θ ) − 1= 2sec 2 θ
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