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July 24, 2014

July 24, 2014

Posted by **Kayla** on Wednesday, September 28, 2011 at 12:56pm.

A. Find the magnitude m/s

B. Find the direction ° (above the horizontal)

- Physics -
**tchrwill**, Wednesday, September 28, 2011 at 1:57pmFor a projectile fired at an angle of µº to the horizontal with a velocity of Vm/s, the maximum height reached derives from h = V^2(sin^2(µ))/29 and the distance traveled parallel to the ground derives from d = V^2(sin(2µ))/g

1--Vcos(µ) = 50/2.75 = 18.181818

2--Vsin(µ) = 9.8(2.75) = 26.95

3--Vsin(µ)/Vcos(µ) = 26.95/18.18 = 1.48239

4--(µ) = arctan(1.48239) = 56º

5--Therefore, the initial launch velocity is V = 18.1818/cos(56) = 32.51m/s in the direction of 56º above the horizontal.

I hope this has been of some help to you.

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