Jane is running from the ivory hunters in the jungle. Cheetahthrows a 7.0-m-long

Question

Jane is running from the ivory hunters in the jungle. Cheetahthrows a 7.0-m-long vine toward her. Jane leaps onto the vine witha speed of 4.0 m/s. When she catches the vine, it makes an angle of20 degrees with respect to the vertical. (a) When Jane is at herlowest point she has moved downward a distance h from the heightwhere she originally caught the vine. Show that h is given byh=L-Lcos20, where L is the length of the vine. (b) How fast is Janemoving when she is at the lowest point in her swing? (c) How highcan Jane swing above the lowest point in her swing? I worked out part a but can't seem to figure out how to work band c. Jane is running from the ivory hunters in the jungle. Cheetahthrows a 7.0-m-long vine toward her. Jane leaps onto the vine witha speed of 4.0 m/s. When she catches the vine, it makes an angle of20 degrees with respect to the vertical. (a) When Jane is at herlowest point she has moved downward a distance h from the heightwhere she originally caught the vine. Show that h is given byh=L-Lcos20, where L is the length of the vine. (b) How fast is Janemoving when she is at the lowest point in her swing? (c) How highcan Jane swing above the lowest point in her swing? I worked out part a but can't seem to figure out how to work band c.

Explanation / Answer

Conservation of Energy: Ei = Ef KEi + Ui + ... = KEf +Uf + ... (1/2)mvi2 + mgh =(1/2)mvf2 + 0 vf2 = vi2 +2gh vlowest pt. = (vi+ 2gh)1/2 = [(4.0m/s)2 +2(9.81m/s2)(7.0m)(1-cos20o)]1/2 =(24.3 m2/s2)1/2 = 4.93m/s Conservation of Energy: Ei = Ef KEi + Ui + ... = KEf +Uf + ... KEi + Ui + ... = KEf +Uf + ... (1/2)mvi2 + 0 = 0 + mgH H = (1/2)mvlowestpt.2/mg =(1/2)(4.93m/s)2/(9.81m/s2) = 4.96 m . .

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