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Projectile Motion/Dropping an object from the sky

  1. Jan 22, 2009 #1
    An eagle is flying horizontally at 7.9 m/s with a fish in its claws. It accidentally drops the fish.

    (a) How much time passes before the fish's speed doubles?

    (b) How much additional time would be required for the fish's speed to double again?


    Initial velocity in the x direction is 7.9m/s. I assumed final velocity in x direction was 7.9m/s and acceleration was 0. But when I use the equations to find one of the other unknown variables everything cancels out and I get zero. There is an example in the book, but it gives the height the fish is dropped from. I feel like I need to know another variable in order to solve this.
     
  2. jcsd
  3. Jan 22, 2009 #2
    Are you sure you're reading the question right? If the fish is in free fall the only acceleration on it is due to gravity. The speed in the x direction will NEVER double unless there is some acceleration in the x direction.

    The distance may double if you assume it can fall long enough, perhaps that is what the question is after.
     
  4. Jan 22, 2009 #3
    That's what I did. I had Ax=0 m/s^2, Vix= 7.9 m/s, Ay= -9.8 and Viy= 0 m/s and when I plug them into any equation it is always zero
     
  5. Jan 22, 2009 #4

    hage567

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    Could you show your calculation?
     
  6. Jan 22, 2009 #5
    I can find either displacement or time. With either one I get 0.

    Vf= Vi + at --> 7.9= 7.9 + (0)t
    Vf^2= Vi^2 + 2ax --> 7.9^2= 7.9^2 + 2(0)x
     
  7. Jan 22, 2009 #6
    What are you finding the displacement of or what would you be finding the time of?
     
  8. Jan 22, 2009 #7
    (a) How much time passes before the fish's speed doubles?

    (b) How much additional time would be required for the fish's speed to double again?
     
  9. Jan 22, 2009 #8
    If the eagle is flying completely horizontal at 7.9m/s when it drops the fish then its speed in the horizontal will remain constant; it will never double. The fish could sail through free space for eternity and it's speed will remain 7.9m/s.

    The question would make more sense if the 7.9m/s was downwards and the eagle let go. Because then you're finding out how long it takes the acceleration due to gravity to speed up the fish from -7.9m/s to -15.8m/s.
     
  10. Jan 22, 2009 #9

    LowlyPion

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    What they are asking in this problem is at what time does the speed double?

    Speed is the |velocity|.

    Hence if Vx is invariant, the magnitude of the velocity vector will be when

    Vy is equal to (√3)*Vx ---> Vx² + ((√3)*Vx )² = (1 + 3)*Vx² = 4*Vx² ---> |New V| = 2*|Vx|

    So all you need to determine is how long until the Vy is (√3)*Vx
     
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