A physics student on a bungey jump drops a watermelon onto somepeople standing b

Question

A physics student on a bungey jump drops a watermelon onto somepeople standing below. if the student was travelling downward at14.05 ms-1 when she dropped the watermelon and it took 1.7s for thewatermelon to hit the people.

i) how high was the bungey when the student dumped thewatermelon?
ii)what speed did the watermelon have just before itsplattered?
iii) the watermelon had a mass of 1.2 kg. what was the totalchange in energy of the watermelon between the time it wasdropped and just before it hit the people? Explain youranswer.

Explanation / Answer

Initial velocity of the watermelon u = 14.05 m/s time t = 1.7 s (1) Height H = ut + (1/2)gt^2               = (14.05)(1.7) + [(0.5)(9.8)(1.7)^2]               = 23.885 + 14.161 = 38.046 m (2) Speed of the watermelon v = ? From kinematic relation                                    v^2 = u^2 + 2gH                                          = 197.4025 + 745.7016                                          = 943.1041                                       v = 30.71 m/s (3) According to law of conservation of energy total energy remains constant.So change in energy = 0 Total change in energy = [ mgH + (1/2)mu^2 ] - (1/2)mv^2                                     = [1.2*9.8*38.046 + 118.4415] - 565.86246                                    = 0   

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