Falling Object. The motion of an object moving vertically through the air is governed by the equation\(\frac{{{{\bf{d}}^{\bf{2}}}{\bf{y}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ = - g - }}\frac{{\bf{g}}}{{{{\bf{V}}^{\bf{2}}}}}\frac{{{\bf{dy}}}}{{{\bf{dt}}}}\left| {\frac{{{\bf{dy}}}}{{{\bf{dt}}}}} \right|\)where y is the upward vertical displacement and V is a constant called the terminal speed. Take \({\bf{g = 32ft/se}}{{\bf{c}}^{\bf{2}}}\)and V = 50 ft/sec. Sketch trajectories in the yv-phase plane for \( - 100 \le {\bf{y}} \le 100, - 100 \le {\bf{v}} \le 100\)starting from y = 0 and y = -75, -50, -25, 0, 25, 50, and 75 ft/sec. Interpret the trajectories physically; why is V called the terminal speed?

Step by step solution

01

Find trajectory

The trajectory approaches the value\(v = - 50\). This means that the object will be falling

through the air at\(50{\rm{ }}ft/sec\). This is the terminal velocity and that velocity is the maximum

velocity attainable by the object as it falls through the air.

02

Step 2:The trajectory

The trajectory can be plotted by mat lab.

Therefore, the terminal velocity is the maximum velocity attainable by an object as it falls through the air.

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