A Doubt from Halliday Resnick Krane

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SUMMARY

The discussion focuses on the behavior of projectile motion under the influence of drag force, as described in Chapter 4 of Halliday Resnick Krane. It establishes that the descending motion of projectiles is significantly steeper than the ascending motion due to the effects of air resistance, particularly quadratic drag. The experiment with paper wads and Styrofoam packing peanuts illustrates the rapid increase in air resistance, which limits the distance a projectile can travel before falling vertically. The mathematical representation of the motion under linear drag is provided, detailing the equations governing velocity and position over time.

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  • Basic knowledge of differential equations
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vibha_ganji
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Hello! In Chapter 4 of Halliday Resnick Krane on Pg. 73, they write “the descending motion is much steeper than the ascending motion” when discussing how the trajectory of a projectile changes when we consider drag force. Why is this statement true?
 
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You could think about terminal velocity in the vertical direction compared to terminal velocity in the horizontal direction.

Edit: Alternately, you could take a bunch of pieces of paper wad them up into a balls and throw them into the air at a 45 degree angle, both softly and as hard as you can. What I would expect you to observe is that there is a fairly sharp cut-off distance beyond which you cannot throw and that the paper wads will reach this distance and then fall more or less vertically to the floor. This is typical behavior for quadratic drag. Air resistance gets very large, very fast. Gravity only becomes significant once the high initial speed has had time to dissipate.

Styrofoam packing "peanuts" work more dramatically than paper wads to demonstrate this behavior.
 
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In the presence of linear drag, the equation of motion is
\begin{align*}
\dot{\mathbf{v}} + 2\gamma \mathbf{v} = \mathbf{g} \implies
\begin{cases}
\dot{v}_x + 2\gamma v_x &= 0 \\
\dot{v}_y + 2\gamma v_y &= -g
\end{cases}
\end{align*}A solution to the homogenous equation is ##e^{-2\gamma t}##. The forced equation for ##v_y## has a particular solution of ##-g/2\gamma##. The velocity is\begin{align*}
\mathbf{v}(t) = \begin{pmatrix} ae^{-2\gamma t} \\ be^{-2\gamma t} - \frac{g}{2\gamma} \end{pmatrix}
\end{align*}If ##\mathbf{v}(0) = (v_{\mathrm{0x}}, v_{\mathrm{0y}})## then ##a = v_{\mathrm{0x}}## and also ##b = g/2\gamma + v_{\mathrm{0y}}##, so that\begin{align*}
\mathbf{v}(t) &= e^{-2\gamma t} \begin{pmatrix} v_{\mathrm{0x}} \\ v_{\mathrm{0y}} + \frac{g}{2\gamma}[1- e^{2\gamma t}] \end{pmatrix}
\end{align*}Integrating, and putting ##\mathbf{x}(0) = 0##, gives\begin{align*}
\mathbf{x}(t) &= \frac{1}{2\gamma} (1- e^{-2\gamma t}) \begin{pmatrix} v_{\mathrm{0x}} \\ v_{\mathrm{0y}} + \frac{g}{2\gamma} \left[ 1 - \frac{2\gamma t}{1- e^{-2\gamma t}} \right] \end{pmatrix}
\end{align*}
 
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