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Kinematics HMWK help

  1. Sep 18, 2006 #1
    Kinematics HMWK help!!!

    hi, im in gr 12 physics and i need help in a kinematics question about Projectile motion... here is the question:

    "a medievil prince is trapped in a castle wraps a message around a rock and throws it from the top of the castlewall with an initial velocity of 12 m/s [42 degrees above the horizontal]. The rock lands just on the far side of the casltes moat, at a level of 9.5 m below the initial level. Determine the (a) time of flight (b) width of moat and (c) velocity at impact"

    i really cant figure this out and i need help ASAP:eek:
     
  2. jcsd
  3. Sep 18, 2006 #2
    Hey, welcome aboard!
    What work have you done on this problem?
    What do you know? What are you looking for? Any ideas of what equation(s) you'll need to use?
     
  4. Sep 18, 2006 #3
    ok well what i have so far is my vertical and horizontal velocities, which are 8.0 m/s and 8.9 m/s respectively....and i dunno where to go from here... i tried using d=v1t + 1/2 at*2 but when i put it in the quadratic formula i cant get it to work :confused: so i dunno where to go from here
     
  5. Sep 18, 2006 #4
    OK, so it sounds like you're solving for x and y components separately, which is very good.

    You know more than that: assign a value for the initial x position (I'd use 0m) and y position (which you could set at zero, but since the rock falls below the throw line, you'd have a lot of negatives flying around. I don't like negatives, so I'd set the initial y position as the height of the wall--which is given). You also know that there is no acceleration in the x direction, and the acceleration in the y direction is acceleration due to gravity.
    I suggest starting by solving for time, using an equation involving y-components only.
     
  6. Sep 18, 2006 #5
    ok ... now i have figured out the time ... it is 2.4 s ... how would i figure out the length of the moat? would i have to use the x components now? and if i do how would i do that?
     
  7. Sep 18, 2006 #6
    yes, yes!

    I'd use [itex]x=x_{0}+v_{0x}t+\frac{1}{2}at^2[/itex]

    Remember that x acceleration = 0.
     
  8. Sep 18, 2006 #7
    ok i got the length of the moat (22 m) !!!! thanx for ur help... but i have one more question that i hope u can answer... how would i figure out the velocity at impact?
     
  9. Sep 18, 2006 #8
    nvm... i got it ... i just used [itex]V_2^2=V_1^2+2ad[/itex]... and i got the answer :) thnx again for ur help
     
  10. Dec 6, 2006 #9
    hey im new here and need some help with a few basic questions
     
  11. Dec 6, 2006 #10
    is anyone online???
     
  12. Dec 6, 2006 #11
    can anyone in here help me with some questions
     
  13. Dec 6, 2006 #12
    it might help if you posted your question...
     
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