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September 2, 2014

September 2, 2014

Posted by **candace** on Sunday, February 14, 2010 at 5:14pm.

R =

vi2sin2θ

g

Use the expression to find the followings.

1. The maximum range of a projectile with launch speed vi.

a)vi^2/2g

b)vi^2/4g

c)4vi^2/g

d)2vi^2/g

e)vi^2/g

(b) The launch angle θ at which the maximum range occurs.

- physics -
**Damon**, Sunday, February 14, 2010 at 5:58pmsin of anything is maximum when anything is 90 degrees or pi/2 radians. That maximum is 1.

Then

R = Vi^2/g times one

and 2 theta = 90 so theta = 45 degrees

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