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Projectile Motion, A study of height and time

  1. Sep 20, 2006 #1
    A Cutnell and Johnson Physics question illustrates a rocket launch:

    A rocket is fired at a speed of 97.0 m/s from ground level, at an angle of 53.0 ° above the horizontal. The rocket is fired toward an 27.0-m high wall, which is located 24.0 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?

    My Calculations:

    Vo = 97.0 m/s
    Theta = 53 degrees
    x = 24 m
    y = 27 m
    Let H illustrate the height of the rocket at point x on the axis.
    t = ?
    a = -9.80 m/s ^2

    Vo sin 53 = Voy = 77.46
    Vo cos 53 = Vox = 58.37

    x = Vox(t)
    t = x/Vox = 24/58.37 = .411 s

    H = Voy(t) + .5 a(t^2)

    Plug and Chug

    H = 77.46(.411) + .5(-9.80)(.411^2) = 31 m

    Distance Rocket clears wall = 31m-27m = 4.00 m <-----Incorrect.

    Is there a problem with my calculations that I am not understanding?
     
  2. jcsd
  3. Sep 20, 2006 #2
    How do u no ur wrong? Looks ok 2 me unless im makin exactly the same mistake. Mayb ur lookin under the wrong chapter at the back of the book 4 ur answers!?
     
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