08_P55InstructorSolution

08_P55InstructorSolution - 8.55. Model: Visualize: Solve:...

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8.55. Model: Assume the particle model and apply the constant-acceleration kinematic equations. Visualize: Solve: (a) Newton’s second law for the projectile is () net wind x x FF m a =− = G x F a m ⇒= where wind F is shortened to F . For the y -motion: 2 1 100 1 0 1 0 2 yy yyvtt a tt =+ −+ ( ) 2 1 01 1 2 0 m sin vt g t θ ⇒=+ 1 0 s t and 0 1 2s i n v t g = Using the above expression for 1 t and defining the range as R we get from the x motion: 2 1 1 0 1 0 2 xx x xvtt a 2 1 10 0 1 1 2 x F x xRv t t m ⎛⎞ ⇒−== +− ⎜⎟ ⎝⎠ 2 00 0 2 sin 2 sin cos 2 vF v v gm g 22 2 2 cos sin sin vv F g θθ We will now maximize R as a function of by setting the derivative equal to 0: 2 cos sin 2sin cos 0 dR v Fv dg m g = 2 0 0 2 cos sin cos2 sin 2 2 Fv g mg v ⇒−== tan2 mg F Thus the angle for maximum range is ( ) 1 1 2 tan / mg F = (b) We have ( ) 2 0.50 kg 9.8 m/s 8.167 0.60 N mg F == ( ) 1 1 2 tan 8.167 41.51 = ° The maximum range without air resistance is i n 4 5 c o s 4 5 R g g °° ′ = = Therefore, we can write the equation for the range
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08_P55InstructorSolution - 8.55. Model: Visualize: Solve:...

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