Max/Min Problem

  • Thread starter Nusc
  • Start date
  • #1
753
2

Homework Statement


Find the angle that gives maximum horizontal distance for a projectile launched from a 9 ft cliff.

[tex]
y = \frac{g}{2V_0^2} sec^2(\theta)x^2 + tan(\theta)x+h
[/tex]
g = -32 ft/s^2
h = 9 ft
V_0 = 26 ft/s

the variable y denotes the vertical position of the object but since we're only interested in the horizontal distance, that is when the object hits the ground, we can assume y=0.

Homework Equations





The Attempt at a Solution


At y = 0 the object is at the ground.
[tex]
0 = \frac{g}{2V_0^2} sec^2(\theta)x^2 + tan(\theta)x+h
[/tex]

Implicit differentiation:
Setting
[tex]
\frac{dy}{d\theta} = \frac{dx}{d\theta} = 0
[/tex]

Yields
[tex] 0=\frac{x sec^2(\theta)(x g tan(\theta) + 1 )}{V_0^2} [/tex]

[tex] tan(\theta) = \frac{-1}{gx}[/tex]

Putting this expression back into
[tex]
0 = \frac{g}{2V_0^2} sec^2(\theta)x^2 + tan(\theta)x+h
[/tex] and using [tex] 1+tan^2(\theta) = sec^2(\theta) [/tex]


we get x = 18.0 ft. which gives 0.09 degrees

The answer is 35 degrees.

How do you do this problem correctly?
 

Answers and Replies

  • #2
35,028
6,774

Homework Statement


Find the angle that gives maximum horizontal distance for a projectile launched from a 9 ft cliff.

[tex]
y = \frac{g}{2V_0^2} sec^2(\theta)x^2 + tan(\theta)x+h
[/tex]
g = -32 ft/s^2
h = 9 ft
V_0 = 26 ft/s

the variable y denotes the vertical position of the object but since we're only interested in the horizontal distance, that is when the object hits the ground, we can assume y=0.

Homework Equations





The Attempt at a Solution


At y = 0 the object is at the ground.
[tex]
0 = \frac{g}{2V_0^2} sec^2(\theta)x^2 + tan(\theta)x+h
[/tex]

Implicit differentiation:
Setting
[tex]
\frac{dy}{d\theta} = \frac{dx}{d\theta} = 0
[/tex]
I don't see any justification for these two equations. The equation above them gives an implicit relationship between x and theta. To maximize x, differentiate with respect to theta, and solve for dx/d(theta), then set it to zero. I haven't worked this through, but that's what I would do.
Yields
[tex] 0=\frac{x sec^2(\theta)(x g tan(\theta) + 1 )}{V_0^2} [/tex]

[tex] tan(\theta) = \frac{-1}{gx}[/tex]

Putting this expression back into
[tex]
0 = \frac{g}{2V_0^2} sec^2(\theta)x^2 + tan(\theta)x+h
[/tex] and using [tex] 1+tan^2(\theta) = sec^2(\theta) [/tex]


we get x = 18.0 ft. which gives 0.09 degrees

The answer is 35 degrees.

How do you do this problem correctly?
 
  • #3
753
2
[tex] \frac{dx}{d\theta}= 0=\frac{x sec^2(\theta)(x g tan(\theta) + 1 )}{V_0^2}
[/tex]

as I've shown above.
 
  • #4
753
2
I got it. I made an algebraic error.
 

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