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Projectile motion finding distance

  1. Mar 15, 2015 #1
    1. The problem statement, all variables and given/known data
    King Arthur's knights use a catapult to launch a rock from their vantage point on top of the castle wall, 12m above the moat. The rock is launched at a speed of 27m/s and an angle of 39∘ above the horizontal. How far from the castle wall does the launched rock hit the ground?

    2. Relevant equations
    sin(theta)
    cos(theta)
    x=1/2at^2

    3. The attempt at a solution
    I found the Vy and Vx components. To find time you see how long it reaches to the peak so I divided Vy/10? I found out time was t=1.69s (probably wrong) Then right there I'm stuck. I don't know what to do with the 12m either. I think I'm on the right track but maybe I'm just wrong.
     
  2. jcsd
  3. Mar 15, 2015 #2
    Do you have an equation for y? That would be the key in incorporating the information about 12m.
     
  4. Mar 15, 2015 #3
    do you mean the y=Vy(t)-1/2at^2? so if that's the case do you plug in 12 into y? and you are solving for t?
     
  5. Mar 15, 2015 #4
    Did you get that equation from your textbook? Doesn't really look correct unless you change what the meaning of y is.

    The way you have your equation, it seems like it should be Δy = Vy(t) - 1/2at^2 or y - y0 = Vy(t) - 1/2at^2, where y is the height at time t and y0 is your initial height.
     
  6. Mar 15, 2015 #5
    I was talking it over with my friend and he gave that equation. I don't understand what you're trying to solve for in that equation.
     
  7. Mar 15, 2015 #6
    Well, you want to know at what time t the rock the hits the ground, as it is a necessary part in finding your final answer: "How far from the castle wall does the launched rock hit the ground?"

    You can't solve for t unless you have an equation for the height of the projectile (y - y0 = Vy(t) - 1/2at^2). If you know your starting height and the height at which you want to find t, you can solve for t using that equation if you keep in mind what y and y0 is in the equation.
     
    Last edited: Mar 16, 2015
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