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October 21, 2014

October 21, 2014

Posted by **nabeelah** on Saturday, September 10, 2011 at 3:34am.

- physics 180 -
**Jai**, Saturday, September 10, 2011 at 3:58amrecall that for a projectile motion, the maximum height that the object can reach is given by the formula,

h,max = [(vo)^2 * sin^2 (θ)]/(2g)

where

vo = initial velocity

θ = angle of release

g = acceleration due to gravity = 9.8 m/s^2

the unknown in the problem is vo. thus we just substitute the given in the formula:

h,max = [(vo)^2 * sin^2 (θ)]/(2g)

1.8 = [(vo)^2 * sin^2 (45)]/(2*9.8)

1.8*2*9.8 = (vo)^2 * (1/2)

34.2 = (vo)^2 * (1/2)

70.56 = (vo)^2

vo = sqrt(70.56)

vo = 8.4 m/s

hope this helps~ :)

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