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March 30, 2015

March 30, 2015

Posted by **Jacqueline** on Friday, December 2, 2011 at 12:24am.

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 11:33amthe 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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