Max Horiz. Distance from 9 ft Cliff: Solving for Angle

[Nusc]

Homework Statement

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

$$y = \frac{g}{2V_0^2} sec^2(\theta)x^2 + tan(\theta)x+h$$
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.
$$0 = \frac{g}{2V_0^2} sec^2(\theta)x^2 + tan(\theta)x+h$$

Implicit differentiation:
Setting
$$\frac{dy}{d\theta} = \frac{dx}{d\theta} = 0$$

Yields
$$0=\frac{x sec^2(\theta)(x g tan(\theta) + 1 )}{V_0^2}$$

$$tan(\theta) = \frac{-1}{gx}$$

Putting this expression back into
$$0 = \frac{g}{2V_0^2} sec^2(\theta)x^2 + tan(\theta)x+h$$ and using $$1+tan^2(\theta) = sec^2(\theta)$$we get x = 18.0 ft. which gives 0.09 degrees

The answer is 35 degrees.

How do you do this problem correctly?

 

[Mark44]

Homework Statement

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

$$y = \frac{g}{2V_0^2} sec^2(\theta)x^2 + tan(\theta)x+h$$
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.
$$0 = \frac{g}{2V_0^2} sec^2(\theta)x^2 + tan(\theta)x+h$$

Implicit differentiation:
Setting
$$\frac{dy}{d\theta} = \frac{dx}{d\theta} = 0$$

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
$$0=\frac{x sec^2(\theta)(x g tan(\theta) + 1 )}{V_0^2}$$

$$tan(\theta) = \frac{-1}{gx}$$

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


we get x = 18.0 ft. which gives 0.09 degrees

The answer is 35 degrees.

How do you do this problem correctly?

 

[Nusc]

$$\frac{dx}{d\theta}= 0=\frac{x sec^2(\theta)(x g tan(\theta) + 1 )}{V_0^2}$$

as I've shown above.

 

[Nusc]

I got it. I made an algebraic error.