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## What shape is this?

I continue the theme of resistive forces inspired by Professor Lewin. I look at throwing a spherical object in air and looking at what trajectory it takes. From high school mathematics, the case of no drag leads to the trajectory of a parabola. Decomposing the problem into horizontal and vertical directions and looking only at magnitudes (assuming object starts at origin of reference frame):

$\begin{array}{lll} s_x& =& v_{0x}t\\s_y&=&v_{0y}t -gt^2/2\end{array}$

Hence.

$s_y =\frac{v_{0y}s_x}{v_{0x}} - \frac{g s_x^2}{2v_{0x}^2}$

The case for air drag is complex. Decomposing the problem again one notes:

• horizontal direction: the object experiences a deceleration secondary to the drag force
• vertical direction: If the object is thrown up then both drag and gravity act to decelerate the object till the speed becomes zero. After this the object is accelerated down by gravity and drag opposes this acceleration. This yields piecewise differential equations.
I used Mathematica to set up and solve the differential equations for horizontal and vertical velocities using the respective components of the initial velocity as the initial conditions. I then integrated these to derive the position of the object. The following graphic simulates the throwing of an object of mass 1 kg and radius 0.2 m at $20\sqrt{2}$ m/s at 45 degrees to the horizontal. The no drag case is shown for comparison. The animation rate is a scaled version of the real values.
Notes
• Note the inherent symmetry of the no drag case. Spatial symmetry about the peak of the throw and time symmetry.
• Note the asymmetry of the drag case. In particular the time to peak (zero vertical speed) is 1.49 seconds and the time to returning to $y=0$ is 3.23 seconds, i.e. the trip is shorter than the trip down. This follows intuitively from the considerations above.