08_P55InstructorSolution

08_P55InstructorSolution - 8.55 Model Visualize Solve...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 F F ma = − = G x F a m = where wind F is shortened to F . For the y -motion: ( ) ( ) 2 1 1 0 0 1 0 1 0 2 y y y y v t t a t t = + + ( ) 2 1 0 1 1 2 0 m 0 m sin v t gt θ = + 1 0 s t = and 0 1 2 sin v t g θ = Using the above expression for 1 t and defining the range as R we get from the x motion: ( ) ( ) 2 1 1 0 0 1 0 1 0 2 x x x x v t t a t t = + + 2 1 1 0 0 1 1 2 x F x x R v t t m = = + ( ) 2 0 0 0 2 sin 2 sin cos 2 v F v v g m g θ θ θ = 2 2 2 0 0 2 2 2 cos sin sin v v F g mg θ θ θ = We will now maximize R as a function of θ by setting the derivative equal to 0: ( ) 2 2 2 2 0 0 2 2 2 cos sin 2sin cos 0 dR v Fv d g mg θ θ θ θ θ = = 2 2 2 0 2 2 0 2 cos sin cos2 sin2 2 Fv g mg v θ θ θ θ = = ⎠⎝ tan2 mg F θ = Thus the angle for maximum range is ( ) 1 1 2 tan / mg F θ = (b) We have ( ) ( ) 2 0.50 kg 9.8 m/s 8.167
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern