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Two questions on kinematics

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Q.1 A ball is thrown with speed v from the edge of acliff of height h. Assume that the ground below the cliff is horizontal. At what inclination angle should it be thrown so that it travels the maximum horizontal distance?

    Q.2 An airplane has a speed of v and a range(out and home) of flight of Rin calm waether. Show that innorth blowing wind ofspeed w, it's range becomes (the eq. given below),in a direction whose true bearing is [tex]\theta[/tex].
    Find the direction in which the range is maximum and the value of the maximum range.

    2. Relevant equations

    Q.1 2v2sin2[tex]\theta[/tex]/g

    Q.2 the given eq:
    R' = R(v2 - w2)/v sqrt( v2 - w2 sin2 [tex]\theta[/tex] )


    3. The attempt at a solution

    Q.1
    I tried to divide it into two parts, the part above and the part below the cliff.
    The Range for the part above the cliff is easy,which is 2v2sin2[tex]\theta[/tex]/g .
    But for the part below it, it's really complicated, I tried to do it but finding the time of flight by solving a quadtratic equation, then multiplying the horizontal component of the velocity to obtain the range.
    When I add up the two parts, I diffrentiate the range with respect to theta, which gave me a very complicated equation that I can't solve for any values of theta.

    Q.2
    I tried to slove it by using relative velocity but I am getiing nowhere near the answer. Is there any other way?
     
  2. jcsd
  3. Sep 19, 2009 #2
    I've worked through part one and you will get a complicated looking equation for [tex]\Theta[/tex]. I got:

    (2(u[tex]\hat{}[/tex]2) +2u)*sin[tex]\hat{}[/tex]2([tex]\Theta[/tex]) + h*sin([tex]\Theta[/tex]) - 2u = 0

    Where u is the initial velocity.

    If you replace sin([tex]\Theta[/tex]) with x you can solve for x just like an ordinary quadratic. Then when you have your value for x, let's say you got x = 5, then you replace x with
    sin([tex]\Theta[/tex]). So you have sin([tex]\Theta[/tex]) = 5. Then just solve for [tex]\Theta[/tex].
     
  4. Sep 19, 2009 #3

    kuruman

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    You need to use the kinematic equations to find the horizontal displacement in terms of the given quantities. Then maximize with respect to the angle of projection.

    Sorry, Kalvarin, but the above equation is dimensionally incorrect. You can't add u2 with u and h together. Also, what happened to g?
     
  5. Sep 21, 2009 #4
    Yep, Kalvarin is wrong, but thanks for trying to help.
    I solved question 1, finally, but I still can't solve question 2.
     
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