The 2 most general solution isequals zero if

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Unformatted text preview: 2 +(v sin θ t − gt2 /2)2 = v 2 t2 − vg sin θ t3 + g 2 t4 /4. (51) 4.13. kx force Trying a solution ofderivative x(t) =and αt in kx) = mx gives αless ± k/m, so The ¨ We want the the form of ( Ae thus 2 to never be = than zero. the 2 most general solution isequals zero if derivative d /dt √ √ k/m t x(= 2v 2 t − 3vg sin θBe− gk/m t . + t 2 + 2 t3 (88) 0 t) = Ae 22 =⇒ otherwise 3vg sin θ t + 2 2 We want A = 0, because 0 = g t −the first term vwould become large for √ large t. √ 1 − k/m t 2 g 2 sin2 = − 8v 2 g 2k /m e− k/m t . So x(t) = Be , =⇒which gives 3vg= xθ .± Hence, v (t) θ −x0 t = 2 B sin 0 9v . (52) 2g Therefore, v (0) = −x0 k/m. 4.14. RopeA solution does not exist for t if the discriminant is less than zero, that is, if on a pulley √ ◦ ◦ Let x sin θthe 2 2/3 =each end is5above if θ is less thanaverage height. .5 , then be < distance ⇒ θ < 70. . So and below the or equal to 70 Then the never decreases during the flight. net force along the rope is σ (2x)g , so F = ma gives 2σ gx = σ Lx. So we essentially ¨ have the Exercise.5◦ . Ifwith k/m → 2g/LatYouangle θ larger than θ0 ,the higher end (b) Let θ0 ≡ 70 4.13 you throw a ball . an should therefore pull then there is a down point a speed v (0) =points)g/Lthe flight where d /dt = 0. This means that at with (actually two x0 2 in . this point the ball is moving in the direction perpendicular to the radial line 4.15. Amplitude from you to the ball. So if this radial line is considered to be the slope of a Taking the derivative to find the max (or min) yields tan ω t = D/C . At this time hill, then the ball at this point has a velocity that is perpendicular to the hill. we have The time-reversed motion of the ball therefore satisfies the setup in Exercise 3.51. C D x(t) = C cos ω t + D sin ω t = C · √ +D· √ = C 2 + D2 . (89) C 2 than Conversely, if you throw a ball C 2 an D2 θ smaller + D2 θ0 , then there doesn’t at + angle exist a point whe...
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