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Posted by on Sunday, November 29, 2009 at 3:31pm.

(Projectile Motion) You are being asked to bombard an enemy fortress that sits on a 40 m tall cliff 100 m from your location. Your cannon has an elevation of 37 degrees. What velocity must your cannonball have to hit the walls of the fortress?

  • college physics - , Sunday, November 29, 2009 at 3:53pm

    First check if it is feasible!
    The fortress is 40 m high and 100 m away.
    The angle θ=tan-1(40,100)=21.8°.
    Since θ < angle of elevation of cannon of 37°, it is feasible.

    Let the muzzle velocity be v0, and the angle of elevation be φ.
    Horizontal component of velocity=v0 cos(φ)
    Vertical component of velocity = v0 sin(φ)

    Let Sh(t)=horizontal distance at time t after firing, and
    Sv(t)=vertical distance at time t.
    g=acceleration due to gravity = 9.8 m/s²

    Then
    Sh(t)=v0 cos(φ) t
    To hit target, we get
    Sh(t)=100 m = v0 cos(φ) t
    t=100/(v0cos(φ)) ....... (1)
    Solve for t in terms of v0 and φ.

    Sv(t) = 40 = v0 sin(φ) t - (1/2)gt²
    Substitute t into above equation to and solve for v0.

    There should be two solutions. Any negative value of v0 should be rejected.
    Each retained solution must be substituted into equation (1). Any solution which results in a negative value of t should also be rejected.
    I get approximately v0=47 m/s and t between 2.5 to 3 seconds.

  • college physics - , Sunday, November 29, 2009 at 4:14pm

    thank you very much.. i understand what you are saying, but can you show me the algebra steps to get the 2 solutions?

    i keep substituting the first equation into the second and try to work it out, but for some reason i am still doing it wrong.

  • college physics - , Sunday, November 29, 2009 at 4:17pm

    It's time you show me your work!
    I will be pleased to help you find the problem if there is any.

  • college physics - , Sunday, November 29, 2009 at 6:51pm

    my variables are a little different than yours, but here is what i have..

    Yf=Vosinθ (tf)-.5g(tf)²

    Yf=Vosinθ(xf)/Vocosθ-.5g(xf/Vocosθ)²

    i'm having trouble isolating Vo. am i supposed to use the quadratic equation? my algebra is not very good.. so if you can point me in the right direction it would help alot.

  • college physics - , Sunday, November 29, 2009 at 8:57pm

    Fortunately there is no quadratic equation to solve, just a square-root.

    Yf=Vosinθ(xf)/Vocosθ-.5g(xf/Vocosθ)²
    Yf- sinθ(xf)/ cosθ=.5g(xf/Vocosθ)²

    Vo²
    =0.5g(xf/cosθ)²/(Yf-(xf)tanθ), or
    Vo=sqrt(0.5g(xf/cosθ)²/(Yf-(xf)tanθ))

    Now take out your calculator and substitute the values. You should get about 46.6 m/s.

  • college physics - , Sunday, November 29, 2009 at 9:30pm

    ahhh i finally got it!!! thank you so much!

  • college physics - , Sunday, November 29, 2009 at 10:08pm

    How much did you get?

  • college physics - , Sunday, November 29, 2009 at 10:09pm

    What velocity did you get?

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