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If you have not already done it, do .11(b) and find the
Chapter 3, Problem 3.13(choose chapter or problem)
If you have not already done it, do 3.11(b) and find the speed v(t) of a rocket accelerating vertically from rest in a gravitational field g. Now integrate v(t) and show that the rocket's height as a function of t is
\(y(t)=v_{\mathrm{ex}} t-\frac{1}{2} g t^{2}-\frac{m v_{\mathrm{ex}}}{k} \ln \left(\frac{m_{\mathrm{o}}}{m}\right)\)
Using the numbers given in 3.7, estimate the space shuttle's height after two minutes.
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QUESTION:
If you have not already done it, do 3.11(b) and find the speed v(t) of a rocket accelerating vertically from rest in a gravitational field g. Now integrate v(t) and show that the rocket's height as a function of t is
\(y(t)=v_{\mathrm{ex}} t-\frac{1}{2} g t^{2}-\frac{m v_{\mathrm{ex}}}{k} \ln \left(\frac{m_{\mathrm{o}}}{m}\right)\)
Using the numbers given in 3.7, estimate the space shuttle's height after two minutes.
ANSWER:Step 1 of 7
The equation to be integrated from the ‘11’ problem is:
\(\begin{aligned} m i v & =-\dot{m} v_{\text {ex }}-m g \\ v+g & =-\frac{m}{m} v_{\text {ex }} \\ \frac{d v}{d t}+g & =-\frac{m}{m} v_{\text {ex }} \\ \int d v+g d t & =-\int \frac{m}{m} v_{\text {ex }} d t \\ v+g t & =-v_{\text {ex }} \ln \left(\frac{m}{m_{0}}\right) \end{aligned}\)
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