Mathematics, 16.03.2020 21:32 NotAPDB1441
When the velocity v of an object is very large, the magnitude of the force due to air resistance is proportional to v squared with the force acting in opposition to the motion of the object. A shell of mass 2 kg is shot upward from the ground with an initial velocity of 600 m/sec. If the magnitude of the force due to air resistance is (0.1)v squared, when will the shell reach its maximum height above the ground? What is the maximum height? Assume the acceleration due to gravity to be 9.81 m divided by s squared.
Answers: 1
![t = -\frac{m}{2b}\cdot \left[\tan^{-1} \left(\frac{v}{\sqrt{\frac{m\cdot g}{b} } }\right) - \tan^{-1} \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } }\right)\right]](/tpl/images/0549/2532/c8501.png)
![\frac{v}{\sqrt{\frac{m\cdot g}{b} } }=\tan \left[-\frac{2\cdot b\cdot t}{m} + \tan^{-1}\left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)\right]](/tpl/images/0549/2532/6637d.png)
![v = \sqrt{\frac{m\cdot g}{b} } \left [\frac{\tan \left(-\frac{2\cdot b \cdot t}{m} \right)+ \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)}{1 - \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)\cdot \tan \left(-\frac{2\cdot b \cdot t}{m} \right) }\right]](/tpl/images/0549/2532/11ae3.png)
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