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Projectile Velocity

  1. Jun 4, 2007 #1
    A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 66.4° above the horizontal. The rocket is fired toward an 11.0 m high wall, which is located 20.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?

    I'm really stuck on this question not positive on my work
    -4.9 t^2 +75 t sin 66.4 +0
    horizontal component 75*cos(66.4) and a vertical component (75*sin(66.4))

    I don't know how you would find time
     
  2. jcsd
  3. Jun 4, 2007 #2

    Doc Al

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    Staff: Mentor

    Consider the horizontal motion. You have the distance and velocity.
     
  4. Jun 4, 2007 #3
    I know it relates to time....but im not sure how to tie the two together
     
  5. Jun 4, 2007 #4

    Doc Al

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    What's the simplest relationship between velocity, distance, and time?

    What's the difference between vertical and horizontal motion for a projectile?
     
  6. Jun 4, 2007 #5
    v=change x/change of time
    the horizontal component doesn't have acceleration?
     
  7. Jun 4, 2007 #6

    Doc Al

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    Exactly right on both. The horizontal component is constant speed motion: Distance = speed X time. What's the horizontal speed and distance traveled?
     
  8. Jun 4, 2007 #7
    Horizontal speed:750m/s
    Distance traveled:
    Distance=speedXtime
    29cos66.4=750m/sXtime<<?
     
  9. Jun 4, 2007 #8

    Doc Al

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    The total velocity is only 75 m/s--find the x-component of that.
    is given--no calculation needed.
     
  10. Jun 4, 2007 #9
    oops i meant 75.0 m/s, would it be 29.0
     
  11. Jun 4, 2007 #10

    Doc Al

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    That's close, but with a caculator you should be more accurate.
    V_x = 75.0 m/s * cos(66.4 degrees) = ?
     
  12. Jun 4, 2007 #11
    30.0262774418
     
  13. Jun 4, 2007 #12
    How does the 11.0 m wall come into play do i have to subtract it from 30.0262
     
  14. Jun 4, 2007 #13
    Not yet, you've so far got the horizontal velocity (30.02). We're on to finding the time it takes to travel 20.0m (horizontally).

    Then we can start on the vertical components.
     
  15. Jun 4, 2007 #14
    so I would just multiply 30.02m/s*20m
     
  16. Jun 4, 2007 #15
    V_y=75m/s*sin(66.4)
    then I use the formula (V_y*t-9.80*t^2)/2
    then subtract that answer from 11m right?<<<am I'm on the right track I just keep getting a very large numbers for answers
     
  17. Jun 4, 2007 #16
    Not quite. Remember the relationship between velocity, time and distance?
     
  18. Jun 4, 2007 #17
    Well i think i messed up on the time it should be .666 i think b/c of the formula v=change of x/change in time?
     
  19. Jun 4, 2007 #18
    so if i calculated v_y right and t right my equation would be 68.727*.666+(-9.80m/s^2*(.666)^2)/(2)
     
  20. Jun 4, 2007 #19
    YES I GOT IT THAT TIME!! so i guess i have learned something today
     
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