At takeoff, his velocity is zero, so there is no centripetal acceleration in the vine direction. The acceleration will be initially in the direction perpendicular to the vine, along his trajectory. The component of his weight in that direction (M g sin 43) will equal M a, so
a = g sin 43
Tarzan, who weighs 880 N, swings from a cliff at the end of a 38.0 m vine that hangs from a high tree limb and initially makes an angle of 43.0° with the vertical. Assume that an x axis extends horizontally away from the cliff edge and a y axis extends upward. Immediately after Tarzan steps off the cliff, the tension in the vine is 610 N. What is the acceleration of Tarzan, as a magnitude and direction, during takeoff?
1 answer
1 answer