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October 24, 2014

October 24, 2014

Posted by **Physics** on Monday, December 8, 2008 at 10:42pm.

Vicosg-((vsin(pheta))/2g)

I think this is what the equatin is when I'm asked to

"Derive a formula for the Range of a projectile based on Vi and the Launch angle, pheta. State when and why is this equation at a maximum."

I'm not exactly sure what it means by maximum or how to derrive the forumula. I think what I wrote above is what i am suppose to eventually get I'm just nore sure how to get that and what it means by maximum

- physics -
**Damon**, Monday, December 8, 2008 at 11:59pmhorizontal problem:

u = Vi cos Theta

range = u * time in air

find the time in air with vertical solution:

Vi sin theta = initial speed up

-g = acceleration

so

v = Vi sin Theta - g t

h = height = Vi sin Theta * t - .5 g t^2

the height is zero when t = 0 and again when it hits ground at the end of flight

0 = t (Vi sin Theta - .5 g t)

so t = time in air = 2Vi sin Theta /g

so

range = Vi cos Theta * (2/g) Vi sin theta

NOW

if you want to know the angle Theta for maximum range

d range/ d Theta = 0 at maximum

0 = Vi cos Theta*(2/g)Vi cos Theta -(2/g) Vi sin theta * Vi sin Theta

or

cos^2 Theta = sin^2 Theta

sin Theta/cos Theta = tan Theta = 1

Theta = 45 degrees

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