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Finding launch angle and velocity

  1. Oct 7, 2015 #1
    1. The problem statement, all variables and given/known data
    You are an invading army who wants to launch a rock over the enemy castle’s wall. The wall is 12 meters high and there is a moat surrounding the wall which forces you to launch from a distance of 17 meters away. What angle and velocity should you shoot at in order to just clear the wall at the top of your projectile’s trajectory?
    2. Relevant equations
    No idea.

    3. The attempt at a solution
    I have read through the relevant chapter in my book twice now trying to find a solution or process to find the angle and velocity for a specific height, but all I can find is referring to maximum distance. I would really appreciate any help even just pointing me in the right direction!

    Thanks so much in advance!
     
  2. jcsd
  3. Oct 7, 2015 #2

    SammyS

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    Hello Forrest. Welcome to PF.

    In the future please show an attempt at a solution or at an understanding.

    What must the vertical component of the (initial) velocity be for a projectile to reach a height of 12 meters?
     
    Last edited: Oct 7, 2015
  4. Oct 7, 2015 #3
    My apologies. Wouldn't the initial vertical velocity have to be about 3.33? I could do the problem given either a velocity or an angle but I am extremely lost on how to do it without both. Thanks!
     
  5. Oct 7, 2015 #4

    SammyS

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    How did you get 3.33 ?
     
  6. Oct 7, 2015 #5
    Whoops i mistyped. I meant 15.33, but that was the lowest value I found that could give a maximum vertical height of 12 meters.
     
  7. Oct 7, 2015 #6

    SammyS

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    Which does not answer the question ... How do you find that velocity?
     
  8. Oct 7, 2015 #7
    I graphed a generic polynomial like Ax^2 +Bx+C, and since I know the A value is -4.9 I plugged that in, and I know that the object is starting from the ground so I made C equal to 0 and I just played with the value of B until I got 12 meters in height. I'm sorry if I'm being difficult, my professor has assigned an entire homework on this type of problem and did not explain at all how to solve anything other than a very basic example.
     
  9. Oct 7, 2015 #8

    SammyS

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    Yes, 15.34 m/s is approximately correct.

    With that as the vertical component of velocity, how much time does it take the projectile to reach that height?
     
  10. Oct 7, 2015 #9
    It would take 1.56 seconds i believe.
     
  11. Oct 7, 2015 #10

    SammyS

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    Yes.

    So that should tell you what horizontal component of velocity is needed to reach the wall (a horizontal distance of 17 meters away) in 1.156 1.56 seconds ?
     
  12. Oct 7, 2015 #11
    The horizontal velocity would need to be 14.71 m/s then?
     
  13. Oct 7, 2015 #12

    SammyS

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    Sorry, I had a typo .

    17 meters in 1.56 seconds
     
  14. Oct 7, 2015 #13
    That would make it 10.897 m/s.
     
  15. Oct 7, 2015 #14

    SammyS

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    Now you have the two components of velocity needed. Right ?
     
  16. Oct 7, 2015 #15
    And now I construct a right triangle to find the complete velocity and solve for theta to give me the angle?
     
  17. Oct 7, 2015 #16

    SammyS

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    Yes.
     
  18. Oct 7, 2015 #17
    Thank you so much for your help! I was having trouble with those first couple steps, I really appreciate you breaking it down for me. You're great! :)
     
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