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  • 1 Post By romsek

Thread: projectile

  1. #1
    Senior Member
    May 2016


    a particle is projected from the ground with velocity 50.96 m/s at an angle tan^-1 5/12 to the horizontal .on its upward path it just passes over a wall 14.7 m high. during its flight it also passes over a wall 18.375 high. show that the second wall must not be less than 23.52 m and not more than 70.56 from the first wall

    so far ive gotten here : initial speed in the y direction =19.6 first wall is 14.7 m high 14.7=19.6t-4.9t^2 when i solve this t=1 t=3 it must be 1. distance in the x is 47.04 when it just passes over the first wall i then tried to find tried to find the time when its travelled 47.04+23.52 but it wouldnt work dont really know what to do next
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  2. #2
    MHF Contributor
    Nov 2013

    Re: projectile

    Solve for the time that the projectile is moving upward and attains a height of 14.7 m.

    Use that time to find the corresponding horizontal position.

    Solve for the two possible times the projectile is at height 18.375m. It's a quadratic, you should be able to solve it.

    Use those two times to compute the two corresponding horizontal positions.

    Subtract off the the horizontal position of height 14.7m found in the 2nd step.
    Thanks from markosheehan
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