Need help deriving the general maximum height equation

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SUMMARY

The discussion focuses on deriving the general maximum height equation for projectile motion, specifically YMAX = (Vm²sin²θ)/(2g). Two methods are presented: a calculus approach using the trajectory equation y = xtanθ - (gx²/2VL²cos²θ) and a non-calculus method involving energy conservation principles. Participants emphasize the importance of clearly showing each step in the derivation process to avoid losing points. The non-calculus method is noted to be simpler for some users.

PREREQUISITES
  • Understanding of projectile motion concepts
  • Familiarity with trigonometric functions and their applications
  • Basic knowledge of calculus, particularly derivatives
  • Experience with energy conservation principles in physics
NEXT STEPS
  • Study the derivation of the trajectory equation in detail
  • Explore energy conservation methods in projectile motion
  • Learn how to apply derivatives to find maximum values in calculus
  • Practice solving projectile motion problems using both calculus and non-calculus methods
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Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators looking for effective teaching methods for deriving motion equations.

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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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