​A projectile is fired at v0 = 381.0 m/s at an angle of θ = 66.9o with respect to the horizontal. As

Physics
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A projectile is fired at v0 = 381.0 m/s at an angle of θ = 66.9o with respect to the horizontal. Assume that air friction will shorten the range by 35.1%. How far will the projectile travel in the horizontal direction, R?

Jul 1st, 2015

Thank you for the opportunity to help you with your question!

for a Projectile
R= [(v^2)/g]* sin2θ

R= [((381)^2)/9.8]* sin2(66.9)

R= 14812.3 * 0.72

R= 10691 m

As air friction shorten the range 35.1% so

effect of air friction = (10691/100)* 35.1 =3753 m

so 

R = 10691 - 3753 

R = 6938 m

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Jul 1st, 2015

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