Need help deriving the general maximum height equation

[arbartz]

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 Y_MAX can be defined or expressed in terms of V_m, 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: Y_MAX=

Homework Equations

Trajectory Equation: y=xtanθ-(gx²/2V_L²cos²θ)

The Attempt at a Solution

After some research I found that the equation should be: Y_MAX=(V_m²sin²θ/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!

 

[mfb]

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?

 

[arbartz]

Ok, so I split it into Vert and Horiz, then ignored the Horiz. I would get:

V_f = -V_msin(θ)
V_i = V_msin(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:
V_f² = V_i² + 2ad
then
(-V_mSin(θ))²=(V_mSin(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?

 

[mfb]

I do not understand what you are doing there.

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

 

[Nickie]

Hello,

Thank you for reaching out for help with deriving the general maximum height equation for projectile motion. This equation is an important tool for calculating the maximum height of a projectile and can be derived using both calculus and non-calculus methods.

For the non-calculus method, we can start by setting up a chart with the variables involved in the equation: YMAX, Vm, g, θ, and a trig function. The first step is to identify the relevant equations that involve these variables. One such equation is the trajectory equation, which is y=xtanθ-(gx^2/2V^2L^2cos^2θ). This equation relates the vertical position (y) of the projectile to its horizontal position (x), initial velocity (V), launch angle (θ), and acceleration due to gravity (g).

Next, we can rearrange this equation to isolate the maximum height (YMAX). This can be done by setting the vertical position (y) equal to zero, as the maximum height is reached when the projectile is at its highest point, and solving for x. This gives us x=(V^2sin2θ)/g. We can substitute this value for x into the trajectory equation, giving us y=(V^2sin2θ)/g*tanθ-(g(V^2sin2θ)/g^2)/2V^2L^2cos^2θ. Simplifying this equation gives us y=(V^2sin2θ)/2g-(V^2sin^2θ)/2g. The second term in this equation can be simplified using the trigonometric identity sin^2θ=1-cos^2θ, giving us y=(V^2sin2θ)/2g-(V^2cos^2θ)/2g. Finally, we can factor out V^2/2g, giving us y=(V^2/2g)(sin2θ-cos^2θ). We can then use the double angle identity sin2θ=2sinθcosθ to simplify this equation further, giving us y=(V^2/2g)(2sinθcosθ-cos^2θ). This can be rewritten as y=(V^2/2g)(cosθ(2sinθ-cosθ)). Finally, we can use the trigonometric identity cosθ=1-sin^2θ to get y=(V^2