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Projectile Motion Cannon Question

  1. Jun 29, 2008 #1
    Im new to the forum, so.. hello everyone!!

    I have a quick question on projectile motion that i cant seem to get...

    A cannon 60m from the base of a 25m high cliff shoots a 15kg projectile at 43 degrees above the horizon...

    (a)- what is the minimum muzzle velocity for the shell to clear to top of the cliff...

    now i know that in the x direction- t= 60m/Vo(cos(43)t
    In the y direction- 25m=Vo(cos(43))t+(.5)(9.8)t^2

    I tried about 20 times to solve for either t or Vo and then plug that into the other equation and solve for one of the variables and then put that value into the next formula and solve for that but i cant seem to get the correct answer, which is apparently Vo= 32.6m/s. Where am i going wrong here?? Thanks!
  2. jcsd
  3. Jun 29, 2008 #2


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    Homework Helper

    I don't know if you're just making typos here, but there shouldn't be a t at the end of your first equation.

    In your second one, that should be sin43, not cos43, since it is the y component of the initial velocity. Also, watch your signs.

    For future reference, there is a Homework Forum where you can post homework problems.
  4. Jun 29, 2008 #3
    Ah! yeah i dont know why i had those extra variables in there... ill be sure to post in the homework help section from now on, thanks! Got the answer by the way.
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