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March 1, 2015

March 1, 2015

Posted by **ashley --- HELP PLEASEE!!!** on Monday, March 7, 2011 at 8:47pm.

- physics -- projectile motion -
**jai**, Monday, March 7, 2011 at 8:59pmrecall that the formula for range is:

R = (vo)^2 * sin (2*theta) / (2g)

where

vo = initial velocity

theta = angle of release

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

now, if we're given the initial velocity, the value of range will depend on theta,, observing this, if we want to have a max Range, we have to find the max value of sin (2*theta),, but the max value of sine is always 1, therefore, equating this to 1:

sin (2*theta) = 1

2*theta = arcsin 1

2*theta = 90 degrees

theta = 45 degrees

hope this helps~ :)

- physics -- projectile motion -
**jai**, Monday, March 7, 2011 at 9:04pmoops, sorry the formula for range must be

R = (vo)^2 * sin (2*theta) / g

i just remembered that the 2g is for the max height:

h,max = (vo)^2 * [sin (theta)]^2 / 2g

but explanation and proving is still the same~ :)

- physics -- projectile motion -
**John**, Monday, August 29, 2011 at 4:46amthe greatest sin value of theta is the sin of 45*

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