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## 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?