Maximizing Range for projectile motion

  • Thread starter jhong213
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  • #1
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I am taking an online intro to physics course and was required to watch lecture videos.

My prof tells me to maximize my range by differentiating the function. My calculus is a bit rusty could someone refreshen my memory of how to do this? I believe i am solving for α.

(dΔx/ dα) = 0 , (-tanα)Δx = ViSinα(Δx / ViCosα) - 1/2g(Δx/ViCosα)^2
 

Answers and Replies

  • #2
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For a mass launched at θ° above the horizontal at an initial speed V, it is easy for you to show that the range R is given by R = VxV /g x sin2θ.

From this you can see intuitively what the angle θ must be for maximum range.
 
  • #3
Ken G
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Yes, some manipulation of your (-tanα)Δx = ViSinα(Δx / ViCosα) - 1/2g(Δx/ViCosα)^2 leads to 1 = -1+ g*Δx/(Vi2*2*Sina*Cosa). I'm not sure what your expression was trying to mean, but if we solve for Δx, and note that 2*Sina*Cosa = Sin(2a), you get daqddyo1's expression for the range, except that it is multiplied by 2. Whichever one is right, it certainly seems that rather than using calculus, it is easier to just solve for Δx algebraically, using trig identities to see where it is maximized. You certainly don't want to take derivatives with respect to a before you have at least divided through the whole expression by tana.
 

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