Physics Honors

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A lovesick lad wants to throw a bag of candy and love notes into the open window of his girlfriend's bedroom 10 m above. Assuming it just reaches the window, he throws the love gifts at 60° to the ground:
a) At what velocity should he throw the bag?
b) How far from the house is he standing when he throws the bag?

  • Physics Honors -

    the equation of motion for an object thrown from (0,0) at an angle θ with velocity v is

    y(x) = -g/(2v2 cos2θ) x2 + xtanθ

    the range (where y=0 again) is

    r = v2 sin2θ/g

    the maximum height reached is

    h = v2 sin2θ/2g

    So, we know that
    h = 10
    θ = 60°

    10 = v2 (3/4)/(2*9.8)
    10 = .038 v2
    v2 = 263.16
    v = 16.22

    The range is twice the distance to the balcony, so the balcony is at half the range:

    r = 16.222 sin(120)/9.8
    = 263.09 * √3/2 / 9.8
    = 23.24

    so, he stood 11.62m from the house

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