# Calculating Maximum Height: Confusion & Solutions

• annamal
In summary, the conversation discusses the calculation of the maximum height required for a projectile to reach the greatest range when shot horizontally at an angle of 0 degrees, ignoring air resistance. The formula for the first derivative of range as a function of height is used to determine this maximum height, which is found to be zero. However, the question is then raised about whether this height truly maximizes the first derivative, as it may not necessarily be the case. It is suggested to plot the first derivative to better understand its behavior.
annamal
Homework Statement
What is the minimum height β0 above ground that is required to generate the greatest range shooting horizontally with an angle of 0 degrees and discounting air resistance? π£0 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:
π¦=β0β0.5ππ‘^2=0ββ0=0.5ππ‘^2βπ‘=sqrt(2*β0/g)

π₯=π£0*π‘β substituting t βπ₯=π£0*sqrt(2*β0/g)

ππ₯/πβ0=π£0/(π*sqrt(2*h0/g))=0 for maximum β0=0.
confused. can someone tell me how I am calculating this wrong?

berkeman
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0 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:
ππ₯/πβ0=π£0/(π*sqrt(2*h0/g))=0 for maximum β0=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:
ππ₯/πβ0=π£0/(π*sqrt(2*h0/g))=0 for maximum β0=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.

Are there any zeroes for the first derivative?

## 1. How do you calculate maximum height?

To calculate maximum height, you can use the formula: h = (v^2 * sin^2ΞΈ) / 2g, where h is the maximum height, v is the initial velocity, ΞΈ is the angle of launch, and g is the acceleration due to gravity.

## 2. What is the most common confusion when calculating maximum height?

The most common confusion when calculating maximum height is not properly understanding the variables in the formula. Many people confuse the initial velocity with the final velocity, or forget to convert the angle to radians.

## 3. How can I avoid making mistakes when calculating maximum height?

To avoid making mistakes, it is important to carefully identify and understand each variable in the formula. Double-check your calculations and make sure you are using the correct units for each variable.

## 4. Can I use the same formula for calculating maximum height in different scenarios?

Yes, the formula for calculating maximum height can be used in different scenarios as long as the variables are appropriately adjusted. For example, if the acceleration due to gravity is different, the value of g will need to be changed in the formula.

## 5. Are there any other factors that can affect the calculation of maximum height?

Yes, there are other factors that can affect the calculation of maximum height, such as air resistance, wind speed, and the shape of the object being launched. These factors may require more complex calculations to accurately determine the maximum height.

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