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?

3 answers

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

eggs

I dont get it :(((