Asked by Jacqueline

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

y(x) = -g/(2v<sup>2</sup> cos<sup>2</sup>θ) x<sup>2</sup> + xtanθ

the range (where y=0 again) is

r = v<sup>2</sup> sin2θ/g

the maximum height reached is

h = v<sup>2</sup> sin<sup>2</sup>θ/2g

So, we know that
h = 10
θ = 60°

10 = v<sup>2</sup> (3/4)/(2*9.8)
10 = .038 v<sup>2</sup>
v<sup>2</sup> = 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<sup>2</sup> sin(120)/9.8
= 263.09 * √3/2 / 9.8
= 23.24

so, he stood 11.62m from the house


Answered by Anonymous
eggs
Answered by Anonymous
I dont get it :(((
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