Time to Peak for Quadratic DragA vertically launched object will be presumed to experience an air resistance force proportional to the square of its speed. For a launch speed of v0 the object is to calculate the time to the peak. The expressions will be developed for the two forms of air drag which will be used for trajectories: but the simpler -cv2 form will be used initially for simplicity and the forms for terminal velocity vt and characteristic time τ will be used. The motion equation for this vertical launch is which can be integrated for time t in the form but this is a non-trivial integral. Using an integral table it can be expressed as We finally reach an expression for the velocity as a function of time: Now to find the time at the peak of the vertical trajectory, we can set the velocity equal to zero, yielding This expression is used in the vertical trajectory calculation. |
Index Fluid friction Reference A. Douglas Davis Sec. 2.6 | ||
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Peak Height for Quadratic DragA vertically launched object will be presumed to experience an air resistance force proportional to the square of its speed. For a launch speed of v0 the object is to calculate the peak height. The expressions will be developed for the two forms of air drag which will be used for trajectories: but the simpler -cv2 form will be used initially for simplicity and the forms for terminal velocity vt and characteristic time τ will be used. Since we have an expression for velocity as a function of time from above, it can be integrated to find an expression for distance y Again, a good integral table is needed to obtain an expression for y: Using the expression for velocity above, it is found that at the peak where v=0: The expression for the peak height is then This expression for ypeak is used in the vertical trajectory calculation. |
Index Fluid friction Reference A. Douglas Davis Sec. 2.6 | ||
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