Chapter 2: Problem 17
Taking Earth's orbit to be a circle of radius \(1.5 \times 10^{8} \mathrm{km},\) determine Earth's orbital speed in (a) meters per second and (b) miles per second.
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Chapter 2: Problem 17
Taking Earth's orbit to be a circle of radius \(1.5 \times 10^{8} \mathrm{km},\) determine Earth's orbital speed in (a) meters per second and (b) miles per second.
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If you know the initial velocity \(v_{0}\) and the initial and final heights \(y_{0}\) and \(y,\) you can use Equation 2.10 to solve for the time \(t\) when the object will be at height \(y .\) But the equation is quadratic in \(t,\) so you'll get two answers. Physically, why is this?
If you travel in a straight line at \(50 \mathrm{km} / \mathrm{h}\) for \(50 \mathrm{km}\) and then at \(100 \mathrm{km} / \mathrm{h}\) for another \(50 \mathrm{km},\) is your average velocity \(75 \mathrm{km} / \mathrm{h} ?\) If not, is it more or less?
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A Frisbee is lodged in a tree 6.5 m above the ground. A rock thrown from below must be going at least \(3 \mathrm{m} / \mathrm{s}\) to dislodge the Frisbee. How fast must such a rock be thrown upward if it leaves the thrower's hand \(1.3 \mathrm{m}\) above the ground?
You're a consultant on a movie set, and the producer wants a car to drop so that it crosses the camera's field of view in time At. The field of view has height \(h .\) Derive an expression for the height above the top of the field of view from which the car should be released.
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