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Urgent Please help me with this 2d Kinematics Problem

  1. Sep 15, 2008 #1
    1. The problem statement, all variables and given/known data
    An owl accidentally drops a mouse it was carrying while flying horizontally at a speed of 5 m/s. The mouse drops to the ground 10m below. What is the mouse's impact velocity? Give the answer as a magnitude and clearly defined angle.

    As of right now I'm not even sure how to approach this problem, if I had an angle to work with I think I could get through it but at the moment I'm bamboozled.

    Thanks for any help :D
  2. jcsd
  3. Sep 15, 2008 #2
    All i know is that you have to use quadratic formula on -4.9t^2+5t+10 to get the time it took to get to the ground. then divide 10m by t to get the speed at impact. no idea how to get the angle. on second thought i think the angle is the slope of the tangent of the parabola at where it hits the x axis on the right side but not sure.
  4. Sep 15, 2008 #3
    this is 2 dimensional though, wouldn't using the quadratic formula like that indicate that the mouse was shot straight up in the air and then is falling back down? I think I have to split it into X and Y components, but I'm not sure how to do so.

    Edit: Duh, vertical motion is independent of horizontal motion, so yea you're right. Still unsure how to get the angle out of this though.
    Last edited: Sep 15, 2008
  5. Sep 15, 2008 #4


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    Hi Istinkatmath,

    The impact velocity is the final velocity (right before it hits ground). How can you find the components of the final velocity? Do you see what to do with them?
  6. Sep 15, 2008 #5
    i think that's my problem at the moment, i don't see what to do with the components.

    Edit: thanks for helping and caring enough to respond

    I solved the quadratic equation and got my final velocity to be 4.933 m/s, now I just have to figure out what to do with it : /
  7. Sep 15, 2008 #6


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    I don't see how you got that result. Can you show the work you did to get it?

    Remember that the horizontal and vertical components are handled separately. So think about what you know for the horizontal motion (initial horizontal velocity, final horizontal velocity, horizontal displacement, horizontal acceleration and time). You're looking for the x-component of the final velocity, so how could you find it?

    Separately, think about the vertical motion and what you know about it in this problem (initial vertical velocity, final vertical velocity, vertical displacement, vertical acceleration and time). With these, you're looking for the y-component of the final velocity.
  8. Sep 16, 2008 #7
    yea i was wrong. so when i am setting up the y(t) function for motion, would it be the following? y= 10 + Voy(t)-1/2gt^2 ? All of the examples in my book give an angle to work with from the beginning, and I don't understand how I can find t without it.
    Last edited: Sep 16, 2008
  9. Sep 16, 2008 #8


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    That's right; the general equation is:

    y=y_0 + v_{0y} t +\frac{1}{2}a_yt^2

    You've already plugged in [itex]y_0=10[/itex] and [itex]a_y=-g[/itex]. You also know (from the problem) what [itex]y[/itex] and [itex]v_{oy}[/itex].

    The value [itex]y[/itex] is the coordinate of the mouse when it hits the ground; if it starts at 10m, what is it at after it falls 10m?

    The value [itex]v_{0y}[/itex] is how much of the initial velocity that is in the vertical direction. What would that be?

    Once you plug those in, your only unknown should be time and you can find it.
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