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Ch 02: Motion Along a Straight Line

Chapter 2, Problem 2

A 15-kg rock is dropped from rest on the earth and reaches the ground in 1.75 s. When it is dropped from the same height on Saturn's satellite Enceladus, the rock reaches the ground in 18.6 s. What is the acceleration due to gravity on Enceladus?

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Lord. So this problem we have a 30 kg brick which is released from a certain height on earth and touches the surface of the ground in 300.217 seconds the same brick touches the surface of the ground in nine seconds when it is released from the same height on MARS MArs is natural satellite photos. What will be the acceleration due to gravity on the surface of photos? So you have some brick, it reaches the ground in free fall on earth. T equals .2177s. We can use t. sub falling some delta. I don't know on phobos, Same rick is falling the same pipe. T. P. Is equal to nine seconds. Now in this case the gravity is acting on the brick on earth. So it's naked 9.81 in three seconds. And in the case of phobos it is some different gravity G. T. And I'm not now we can start off by solving this problem by recalling United Nations, the F. B. I was, we have squared it was the I. Square was to a double X. And the packs includes the I. T. Was gonna have a. T. Square. Now we have tea, we have G. For Earth and we're looking for the G. Of bobo's. We have some common height. The only equation which includes these terms this equation three includes time acceleration due to gravity on both planets. And we know that in both cases the I. Is equal to zero m per second that is not we have delta H. Which is the same in both cases. So we can set up equation three for both cases where delta H. Is equal to the I. T. Plus one half G. Of earth. T. Sweat where this term is just equal to zero as V. I. Is equal to zero. We get that delta H. For the Earth case and just want to have G. Of earth honesty of Earth split. So this is the earth case. And for the phobos case we have the same delta age which is equal to B. I. T. Phobos plus one a half G phobos team was We know that this is also zero. So we get to the delta H. Is equal to want to have G phobos, T phobos swear this is phobos case. Now these two equations are equal because the velvet ages are equal so we can set one a half G. Of Earth times T. Of Earth squared equal to one half the provost T. Of phobos squared. Now that this won't have canceled. We can now directly solve the gravity of photos. It's G 10 million square over T. P. Screen and now we know G. We know tv. We know T. P. So we can substitute the values G. E. It's -921 meters per second squared T. V 0.27 seconds squared. We have TVs nine seconds it was explosive. We get that G. P Is equal to 0.0057 m/s squared. And some choice indeed, on this house have a great day.
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High-speed motion pictures (3500 frames/second) of a jumping, 210–μg flea yielded the data used to plot the graph in Fig. E2.54. (See 'The Flying Leap of the Flea' by M. Rothschild, Y. Schlein, K. Parker, C. Neville, and S. Sternberg in the November 1973 Scientific American.) This flea was about 2 mm long and jumped at a nearly vertical takeoff angle. Use the graph to answer these questions: (c) Find the flea's acceleration at 0.5 ms, 1.0 ms, and 1.5 ms.

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