Calculating Maximum Height: Confusion & Solutions

AI Thread Summary
The discussion revolves around calculating the maximum height required to achieve the greatest horizontal range when shooting at an angle of 0 degrees, while neglecting air resistance. The initial calculations suggest that the maximum height, h0, approaches zero, leading to confusion about the validity of this conclusion. Participants highlight that setting the first derivative of the range function to zero does not necessarily indicate a maximum height, as it may simply indicate a point where the derivative is zero. The need for a clearer understanding of the relationship between height and range is emphasized, suggesting that plotting the first derivative could provide better insight. Overall, the conversation reflects a struggle to reconcile mathematical results with physical intuition regarding projectile motion.
annamal
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Homework Statement
What is the minimum height β„Ž0 above ground that is required to generate the greatest range shooting horizontally with an angle of 0 degrees and discounting air resistance? 𝑣0 is initial velocity.
Relevant Equations
y = h + vt + 0.5a*t^2
vx = v0*cos(theta)
vy = v0*sin(theta) - g*t
I am calculating it like this:
𝑦=β„Ž0βˆ’0.5𝑔𝑑^2=0β†’β„Ž0=0.5𝑔𝑑^2→𝑑=sqrt(2*β„Ž0/g)

π‘₯=𝑣0*𝑑→ substituting t β†’π‘₯=𝑣0*sqrt(2*β„Ž0/g)

𝑑π‘₯/π‘‘β„Ž0=𝑣0/(𝑔*sqrt(2*h0/g))=0 for maximum β„Ž0=0.
confused. can someone tell me how I am calculating this wrong?
 
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The question doesn't seem to make much sense.

eg: "angle of 0 degrees".
 
Last edited:
annamal said:
Homework Statement:: What is the minimum height β„Ž0 above ground that is required to generate the greatest range shooting horizontally with an angle of 0 degrees and discounting air resistance?
Uh ... as high as you can get. You don't see how that question makes no sense?
 
annamal said:
𝑑π‘₯/π‘‘β„Ž0=𝑣0/(𝑔*sqrt(2*h0/g))=0 for maximum β„Ž0=0.
confused. can someone tell me how I am calculating this wrong?
Does ##dx/dh_0=0## necessarily give you the point where x reaches a maximum? In what two ways might that not be true.
 
annamal said:
𝑑π‘₯/π‘‘β„Ž0=𝑣0/(𝑔*sqrt(2*h0/g))=0 for maximum β„Ž0=0.
So you have this formula for the first derivative of range as a function of height. And you seem have observed that this first derivative is maximized when height ##h_0## approaches zero.

But you seem to have lost track of what you were doing. You are looking for a height that makes the first derivative zero. Not a height that maximizes it.

If you plotted the first derivative, you could see that it looks a bit like just the first quadrant of a hyperbola.

[Googled up stock hyperbola]
1648469511641.png


Are there any zeroes for the first derivative?
 
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