Posted by Jacqueline on Friday, December 2, 2011 at 12:24am.
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  Steve, Friday, December 2, 2011 at 11:33am
the equation of motion for an object thrown from (0,0) at an angle θ with velocity v is
y(x) = g/(2v^{2} cos^{2}θ) x^{2} + xtanθ
the range (where y=0 again) is
r = v^{2} sin2θ/g
the maximum height reached is
h = v^{2} sin^{2}θ/2g
So, we know that
h = 10
θ = 60°
10 = v^{2} (3/4)/(2*9.8)
10 = .038 v^{2}
v^{2} = 263.16
v = 16.22
The range is twice the distance to the balcony, so the balcony is at half the range:
r = 16.22^{2} sin(120)/9.8
= 263.09 * √3/2 / 9.8
= 23.24
so, he stood 11.62m from the house
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