1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Shooting a cannonball over a cliff

  1. Jul 4, 2009 #1
    1. The problem statement, all variables and given/known data

    A cannon 60m horizontally from the base of a 25m cliff is fired with a speed of 32.6m/s at an angle of 43 degrees above the horizontal towards the cliff. How far does the shell land past the edge of the cliff assuming its elevation is constant

    2. Relevant equations

    y = v0yt -.5gt^2

    x= v0xt

    3. The attempt at a solution

    I plug in the to the first equation

    25m = 32.6sin(43)t -4.9t^2

    and the quadratic formula gives me two answers: 2.46s and 2.06s

    so I plug this into the second equation giving me distances of 49.0m and 58.6

    which seems to mean that the cannonball does not actually clear the cliff...but my online how system is asking for an answer.
     
  2. jcsd
  3. Jul 4, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    If it gives 2 answers, isn't one when it passes the height going up and the other when it lands?. So don't you first need to figure what the horizontal velocity is? Then take the longer time and figure the horizontal distance from the gun (2.46*32.6*cos(43)) then subtract the distance from the base of the cliff?

    Edit: I see. Your solutions are not reaching the cliff. Check your quadratic again or maybe enter -1.4 m then
     
  4. Jul 4, 2009 #3

    LowlyPion

    User Avatar
    Homework Helper

    I get 2.44 sec and 2.09 sec. Which falls even a little shorter.

    Are you sure you have the right statement of the problem?

    As stated, I would have to wonder.
     
  5. Jul 4, 2009 #4
    The way I read this, the cannon is sitting at the bottom of a valley shooting a cannon ball towards a cliff that is 60m away and 25m high.

    Therefore, shouldn't you be looking to see where the canon is vertically at t where xhorizontal = 60m?

    I get about 24.887m at t = 2.51655 for xvertical. Assuming this is just a rounding error, the shell would land 0m past the cliff's edge.
     
  6. Jul 4, 2009 #5

    negitron

    User Avatar
    Science Advisor

    That's going to happen at two points; one on the way up and one on the way back down. You found the first.
     
  7. Jul 4, 2009 #6

    ideasrule

    User Avatar
    Homework Helper

    In other words, it's true that the cannon ball just barely avoids hitting the cliff, but it's still moving after scraping by and will eventually hit the ground.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Shooting a cannonball over a cliff
  1. Shooting over a Hill (Replies: 3)

Loading...