Physics

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For a science fair competition, a group of High School students built a kicker-machine that can launch a golf ball from the origin with a velocity of 12.3 m/s and initial angle of 29.9° with respect to the horizontal.

*Where will the golf ball fall back to the ground?

*How High will it be at it's highest point of trajectory?

*What is the ball's velocity vector (in Cartesian components) at the highest point of its trajectory? x, then y

*What is the ball's acceleration vector (in Cartesian components) at the highest point of its trajectory? x, then y

  • Physics -

    Answer the last two questions immediately.
    At the highest point the ball has no y velocity and the x velocity is the same for the whole trip, u = 12.3 cos 29.9
    During the whole trip the acceleration is only in the y direction and is -9.8m/s^2

    Now do the problem.
    Y(vertical) problem first.
    Vi = 12.3 sin 29.9
    at top v = 0
    0 = Vi - 9.8 t
    9.8 t = 12.3 sin 29.9
    solve for t at top, which is half the trip time
    h = Vi t - 4.9 t^2
    solve for h, height at top
    T = 2 t, total time in air
    then d = u T = [12.3 cos 29.9 ] * T

  • Physics -

    thanks a lot@Damon

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