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Kinematics in 2D ?

  1. Sep 20, 2004 #1
    Kinematics in 2D ????

    i have a few question can someone help me...thanks in advanced

    1. An airplane with a speed of 97.5m/s is climbing upward at an angle of 50.0 degree with respect to the horizontal. When the plane's altitude is 732m, the pilot releases a package. (a) calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) relative to the ground, determine the angle of the velocity vector of the package just before impact.

    2.Stones are thrown horizontally with the same velocity from the tops of 2 different buildings. One stone lands twice as far from the base of the building from which it was thrown as does the other stone. Find the ratio of the height of the taller building to the height of the shorter building.

    3. The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his head, forcing him to move quickly away from the net. Suppose that you loft with an initial speed of 15.0m/s, at an angle of 50.0 degree above horizontal. At this instant your opponent is 10.0m away from the ball. He begins moving away from you 0.30 sec later, hoping to reach the ball and hit it back at the moment that it is 2.10m above its launch point. With what minimum average speed must he move?? ( ignore the fact that he can stretch, so that his racket can reach the ball before he does)
     
  2. jcsd
  3. Sep 20, 2004 #2
    To help you with number one, I would recommend drawing a picture and putting what you know. If you use what you have with the equations found in your book (constant acceleration formulas...) then you can probably figure it out. Also, think about what the path of the objective will be when it is released....
     
  4. Sep 20, 2004 #3
    The initial velocity of the package is the same with the velocity of the plane at the instant it is released.
    [tex]
    \begin{multline*}
    \begin{split}
    &Consider\ the\ vertical\ y\ component:\\
    &Use\ s=ut+\frac{1}{2}at^2\ to\ find\ t\ when\ it\ reaches\ ground.\\
    &Use\ the\ t\ to\ find\ the\ horizontal\ distance\ by\ using\ the\ same\ formula\ but\ now\ a=0.\\
    &Use\ \vec{v}=\vec{u}+\vec{a}t\ to\ find\ the\ velocity\ in\ term\ of\ \vec{i}\ and\ \vec{j}.\\
    \end{split}
    \end{multline*}
    [/tex]
     

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    Last edited: Sep 20, 2004
  5. Sep 20, 2004 #4
    [tex]
    \begin{multline*}
    \begin{split}
    &Use\ \vec{s}=\vec{s_o}+\vec{u}t+\frac{1}{2}\vec{a}t^2\\
    &Consider\ the\ shorter\ building:\\
    &s\vec{i}=h\vec{j}+ut_1\vec{i}-\frac{g}{2}t^2_1\vec{j}\\
    &where\ t_1\ is\ the\ time\ to\ reach\ the\ ground\\
    &Consider\ the\ taller\ building:\\
    &2s\vec{i}=H\vec{j}+ut_2\vec{i}-\frac{g}{2}t^2_2\vec{j}\\
    &where\ t_2\ is\ the\ time\ to\ reach\ the\ ground\\
    \end{split}
    \end{multline*}
    [/tex]
    Solve the equations.
     

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