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A cannon ball is fired at an angle of 30° above horizontal with an initial speed vi = 55 m/s. Assume the cannon ball leaves the cannon at ground level and ignore air resistance.

(a) How long is the cannon ball in the air?
(b) How far from the base of the cannon does the cannon ball hit the ground?
(c) Determine the correct answer. The speed of the stone
is greatest just after it is launched and just before it hits the ground.
is greatest just after it is launched.
is greatest at its maximum height.
is constant throughout its flight.

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1 answer
  1. Vo = 55m/s[30o]
    Xo = 55*Cos30 = 47.63 m/s
    Yo = 55*sin30 = 27.5 m/s.

    a. Y = Yo + g*Tr = 0 @ max ht.
    Tr=-Yo/g = -27.5/-9.8 = 2.806 s.=Rise
    time.

    Tf = Tr = 2.806 s. = Fall time.

    Tr+Tf = 2.806 + 2.806 = 5.61 = Time in
    air.

    b. Dx = Xo*(Tr+Tf) = 47.63m/s * 5.61s =
    267.3 m. from base of cannon.

    c. The first choice is correct.

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