Need help deriving the general maximum height equation

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Homework Help Overview

The discussion revolves around deriving the general maximum height equation for projectile motion. The original poster seeks assistance in understanding how to express the maximum height in terms of initial velocity, gravitational acceleration, launch angle, and trigonometric functions. They mention both calculus and non-calculus methods for the derivation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different methods for deriving the maximum height, including energy conservation and calculus approaches. The original poster expresses uncertainty about starting the derivation and questions the validity of their setup. Others suggest focusing on vertical motion and calculating time to the highest point.

Discussion Status

The discussion is ongoing, with participants offering various approaches and questioning each other's methods. Some guidance has been provided regarding the separation of motion components and the use of energy conservation principles, but no consensus or clear resolution has been reached yet.

Contextual Notes

The original poster is required to show every step in their derivation process, and there is an emphasis on clarity to avoid losing points. There is also a mention of different derivation methods, which may influence the approach taken by participants.

arbartz
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Homework Statement


You need to derive the general maximum height equation for projectile motion. This equation allows you to calculate the maximum height a projectile reaches or could reach during its motion. Like other such equations we derived in this unit, this one is also very short and simple. In this particular equation max height or YMAX can be defined or expressed in terms of Vm, g, θ, and one trig function. there is both a calculus and non-calculus derivation of this particular formula and both require careful and creative thinking.
-The calculus method begins with using the trajectory equation.
-The non-calculus method begins with setting up the charts.

You may choose either method, but you must clearly show every step in the derivation process. Missing or unclear steps will result in a loss of points. Finally clearly write and circle your resulting formula as: YMAX=

Homework Equations


Trajectory Equation: y=xtanθ-(gx2/2VL2cos2θ)


The Attempt at a Solution


After some research I found that the equation should be: YMAX=(Vm2sin2θ/2g)

I really don't even know where to start on this. He said that the non-calculus method is much easier. Any help would be greatly appreciated!
 
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You can split the movement in horizontal and vertical motion and ignore the horizontal part.

Energy conservation is one way to get the maximal height, equations for a free fall will work, too.
In terms of the calculus method: Try to find the derivative of y with respect to x. Which value do you expect at the highest point?
 
Ok, so I split it into Vert and Horiz, then ignored the Horiz. I would get:

Vf = -Vmsin(θ)
Vi = Vmsin(θ)
d = y
t = t
a = g

Is that correct for the chart?

Then I would just have to solve for y and ignore time, using this equation:
Vf2 = Vi2 + 2ad
then
(-VmSin(θ))2=(VmSin(θ))2+2(g)y
but then I run into the problem where y=0. Which is certainly not what I need.

What am I doing wrong?
 
I do not understand what you are doing there.

You could try to calculate the time from launch to highest point, for example.
 

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