physics

a contestant projects a coin with a speed of 7 m/s at an angleof 60 degrees to the horizontal. When the coin leaves his hand, the horizontal distance between the coin and the dish is 2.8m. The coin lands in the dish. calculate the horizontal component of the initial velocity of the coin?

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asked by jt
  1. u = 7 cos 60 the whole time
    so u = 7(1/2) = 3.5 m/s

    end of problem

    If there are more parts to this problem:

    Vi = 7 sin 60 at the beginning

    horizontal problem:
    x = u t
    2.8 = 7 (1/2) t
    t = . 8 seconds in the air

    t at top = .4 s

    h = Vi (.4) - (9.81/2)(.4^2)

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    posted by Damon

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