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Physics / Chapter 6: Force and Motion - II / Section 6-3

The Drag Force and Terminal Speed — Question 9

Section 6-3 · Physics pages 828–829

Question 9 The drag sign changes across an upward launch Take g = 10 m/s2 and neglect buoyancy unless explicitly specified. A 1 kg body is launched vertically upward at 10 m/s in still air. With upward v positive, drag is Fd = −cv|v|, with c = 0.10 kg/m. Only weight and drag act. ascent drag descent drag Motion arrows and drag arrows oppose each other on both branches. (a) Find acceleration at launch and at the apex. (b) Find acceleration when it later descends at 5 m/s. (c) Write the terminal velocity and explain why Fd = −cv2 cannot be used for both ascent and descent.
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Question 9 - Solution The drag sign changes across an upward launch ascent drag descent drag (a) Use the ascent sign. The signed force law gives a(v) = −g −c mv|v|. a(+10) = −10 −.10(10)(10) = −20 m/s2 , a(0) = −10 m/s2 . At the apex, drag vanishes but weight remains. (b) Reverse drag during descent. During descent v = −5 m/s, so v|v| = −25 m2/s2: Fd = −.10(−25) = +2.5 N, a(−5) = −10 + 2.5 = −7.5 m/s2 . (c) Balance only on the descending branch. On the descending branch, v|v| = −v2 and 0 = −g + c mv2 t =⇒vt = − √mg c = −10 m/s . The unsigned-square expression −cv2 always points down and would incorrectly reinforce gravity during descent. The factor v|v| reverses drag when velocity changes sign.

Question 9

The drag sign changes across an upward launch

Take g=10g=10 m/s2^2 and neglect buoyancy unless explicitly specified. A 1 kg body is launched vertically upward at 10 m/s in still air. With upward v positive, drag is Fd=−cv|v|F_d=-cv|v|, with c=0.10c=0.10 kg/m. Only weight and drag act.

Diagram shown in the original SVG worksheet.

  1. Find acceleration at launch and at the apex.

  2. Find acceleration when it later descends at 5 m/s.

  3. Write the terminal velocity and explain why Fd=−cv2F_d=-cv^2 cannot be used for both ascent and descent.

Question 9 - Solution

The drag sign changes across an upward launch

Diagram shown in the original SVG worksheet.

(a) Use the ascent sign.

The signed force law gives a(v)=−g−cmv|v|.a(v)=-g-\frac cmv|v|. a(+10)=−10−.10(10)(10)=−20m/s2,a(0)=−10m/s2.\boxed{a(+10)=-10-.10(10)(10)=-20\ \mathrm{m/s^2}},\qquad \boxed{a(0)=-10\ \mathrm{m/s^2}}. At the apex, drag vanishes but weight remains.

(b) Reverse drag during descent.

During descent v=−5v=-5 m/s, so v|v|=−25v|v|=-25 m2^2/s2^2: Fd=−.10(−25)=+2.5N,a(−5)=−10+2.5=−7.5m/s2.F_d=-.10(-25)=+2.5\ \mathrm N,\qquad \boxed{a(-5)=-10+2.5=-7.5\ \mathrm{m/s^2}}.

(c) Balance only on the descending branch.

On the descending branch, v|v|=−v2v|v|=-v^2 and 0=−g+cmvt2⇒vt=−mgc=−10m/s.0=-g+\frac cmv_t^2\Longrightarrow\boxed{v_t=-\sqrt{\frac{mg}{c}}=-10\ \mathrm{m/s}}. The unsigned-square expression −cv2-cv^2 always points down and would incorrectly reinforce gravity during descent. The factor v|v|v|v| reverses drag when velocity changes sign.

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