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Projectile thrown horizontally off a cliff

  1. Oct 11, 2015 #1
    • Member warned to retain and use the formatting template
    A ball is thrown horizontally from a height of 20m and hits the ground with a speed that is three times its initial speed. What is the initial speed?


    d=vit+1/2at^2


    I found the time of flight vertically first, which is 2.02 sec. And then I am trying to find the initial speed. Even if it is thrown horizontally off a cliff, is the acceleration still 0?
     
  2. jcsd
  3. Oct 11, 2015 #2

    TSny

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    Did you mean to say 0 here? Once the ball leaves your hand, gravity takes over. So, what is the acceleration (magnitude and direction) while the ball is in flight?

    Your value for the time of flight looks correct.
     
  4. Oct 11, 2015 #3
    Doesnt it depend? Horizontally it does not have a acceleration. But does it apply when it is thrown off a cliff? Because it is thrown off the cliff horizontally not straight down.
     
  5. Oct 11, 2015 #4

    TSny

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    No matter how you throw the ball, once it's in flight the only thing "acting on it" is gravity (neglecting air resistance, of course). The acceleration caused by gravity is g downward.
     
  6. Oct 11, 2015 #5
    I am confused now but I was told that horizontally an object has no acceleration?
     
  7. Oct 11, 2015 #6

    TSny

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    Acceleration is a vector quantity. For a projectile, the horizontal component of the acceleration is zero. But the vertical component is 9.8 m/s2 downward.
     
  8. Oct 11, 2015 #7
    Can I solve this problem horizontally? I tried it but it seems a little off.

    Vf=Vi+at
    A would be cancelled because of horizontally
    Vf=Vi?
    I looked in the answer key and they did it differently but my professor taught us this way to find vertically and horizontally. I don't know if it applies to this problem though.
     
  9. Oct 11, 2015 #8

    TSny

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    Yes, you should think about the horizontal and vertical components of the motion.

    Suppose Vo is the initial speed that the object is thrown horizontally. Since the horizontal component of acceleration is zero, what can you say about the value of the horizontal component of velocity just as the ball reaches the ground?
     
  10. Oct 11, 2015 #9
    It's the same as the initial?
     
  11. Oct 11, 2015 #10

    TSny

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    Yes. Now think about the vertical component of motion. What is the initial value of the vertical component of velocity? Can you see a way to get the final vertical component of velocity using the height that the ball was thrown from?
     
  12. Oct 12, 2015 #11
    I got Vf for the vertical component as 19.8. How is this related to the Vf of the horizontal?
     
  13. Oct 12, 2015 #12

    CWatters

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    Your post #9 is correct. The horizontal component is unchanged.
     
  14. Oct 13, 2015 #13
    Bump
     
  15. Oct 13, 2015 #14

    TSny

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    I'm not sure I understand your question here. But, how do you construct the speed of a particle from it's horizontal and vertical components?
     
  16. Oct 13, 2015 #15
    You would just use

    [tex]\Delta \ x = \ v_0 t + \frac{1}{2} \alpha t^2[/tex]

    for both horizontal and vertical
     
  17. Oct 13, 2015 #16

    CWatters

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    You just add (vector add) the horizontal and vertical components to get the impact speed.
     
  18. Oct 13, 2015 #17

    TSny

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    Let's make sure we are together so far. Let Vo be the initial speed of the ball.

    When the ball reaches the ground, how would you express the horizontal and vertical components of the velocity?
     
  19. Oct 13, 2015 #18
    Vertically the final velocity would be 39.2. I don't know how to find it horizontally.
     
  20. Oct 13, 2015 #19

    TSny

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    In post #11 you stated that the vertical component would be 19.8 m/s. And in post #9 you gave the correct answer for the horizontal component of the final velocity.
     
  21. Oct 13, 2015 #20
    The 19.8 is the final velocity of the vertical.

    Vf=Vi only in the horizontal component not the vertical component.
     
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