Recent content by aviodont

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    I used to be good at phyics, but

    When Vy=0, Y is at its highest. So,...
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    I used to be good at phyics, but

    Rifle muzzle at 5ft shoots up at 45 degrees at 3000ft/sec (no wind, etc.): -5=0+2121ft/sec*T-15.16ft/sec^2*T^2 T=139 sec X=0+2121ft/sec*139sec=294819ft But, I never figured out how to get Ymax? (and it's time for a lunch break... all this math is making me crazy!)
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    I used to be good at phyics, but

    Thanks! I am going to try it with a different set of numbers (I might be back!)
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    I used to be good at phyics, but

    zero! :blushing: x = x_0 + v_0 t + (1/2) a t^2 x=0 + 353.56m/sec*72.09sec+ zero*(72.16sec)^2 x=25488.14m+0=25.488km=seems a little far to me still
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    I used to be good at phyics, but

    Not anymore! You need an emoticon for "my head is about to burst!" Well, going online, I found the solutions to be: -.005653 (disregard) 72.16 So T=72.16 x = x_0 + v_0 t + (1/2) a t^2 x=0 + 353.56m/sec*72.16sec+ 4.9m/sec^2*(72.16sec)^2 x=25512.89m+25514.62m=51027.51m=51.027km=no...
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    I used to be good at phyics, but

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    I used to be good at phyics, but

    -2=0+353.56+0.5(-9.8)T^2 -4.9T^2=-355.56 T^2=+72.56 T=8.52 sec x = x_0 + v_0 t + (1/2) a t^2 x=0 + 353.56m/sec*8.52sec + 4.9m/sec^2*72.56sec^2 x=3012.33m+355.54m x=3367.87m The Ymax I thought I would figure by dividing the time in half, but being a perfectionist, how would I account...
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    I used to be good at phyics, but

    No, I cannot, because, although I know Xo, Vo, and A, there are 2 unknowns in that equation: X and T.:cry:
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    I used to be good at phyics, but

    The final vertical displacement would be -2meters. Am I missing something? I have no idea ho high it went nor the time in flight (both of which would be related to 9.8m/s^2) nor the distance traveled (which would be dependent on the time in flight). What am I mssing here?
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    I used to be good at phyics, but

    Yes, I am trying to find the range, which would be x. The original x=0 (the muzzle of the cannon), the original y=2 (height of muzzle... i used 5 feet, let's just call it 2 meters). The final x (the displacement) I do not know.
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    I used to be good at phyics, but

    I do not know the time. Or the initial and final displacments.
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    I used to be good at phyics, but

    The first four I understand. The fifth, x = x_0 + v_0 t + (1/2) a t^2 I believe says that the x-coordinate will be the original starting point (x=0 in this case) plus the original velocity multiplied by time plus 1/2 acceleration multiplied by the time squared. In this case, I could...
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    I used to be good at phyics, but

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    I used to be good at phyics, but

    So we start off with an initial x and y velocity of ~353m/sec; the x component will stay that throughout the time in flight (since we will be ignoring wind resistance and other effects). The Y velocity will start to bleed off at a rate to be determined somewhow by that 9.8 m/s^2 thing, but I...
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    I used to be good at phyics, but

    Arrrgh! You are going to make me think? Oh, well! It is either the sin or cos of 45. Maybe it's pythagorean's theorem? Is it 353.56 for initial x and y? (Honestly, I used to get 100% on my physics tests!)