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## Homework Statement

A package is dropped from the plane which is flying with a constant horizontal velocity of v

_{a}=150 ft/s. Determine the normal and tangential components of acceleration, and the radius of curvature of the path of motion (a) at the moment the package is released at A, (b) just before the package strikes the ground at B.

## Homework Equations

[tex]s = s_{o} + v_{o}t + (1/2)(a_{t})_{c}t^{2}[/tex]

[tex]v = v_{o} + (a_{t})_{c} t[/tex]

[tex]v^{2} = (vo)^{2} + 2(a_{t})_{c} (s - s_{o})[/tex]

[tex]\rho = \frac{1 + ((\frac{dy}{dx})^{2})^{3/2}}{|d^{2}y/dx^{2}|}[/tex]

## The Attempt at a Solution

[tex]s = s_{o} + v_{o}t + (1/2)(a_{t})_{c}t^{2}[/tex]

[tex]x = 0 + (150ft/s)t + 0[/tex]

[tex]x =150t[/tex]

[tex]t = \frac{x}{150}[/tex]

[tex]s = s_{o} + v_{o}t + (1/2)(a_{t})_{c}t^{2}[/tex]

[tex]y = 0 + 0 + (1/2)(-32.2ft/s^{2})t^{2}[/tex]

[tex]y = (-16.1ft/s^{2})t^{2}[/tex]

[tex]y = (-16.1ft/s^{2})(\frac{x}{150})^{2}[/tex]

[tex]y = -0.000716x^{2}[/tex] This is the equation of the path.

[tex]dy/dx = -0.00143x[/tex]

[tex]d^{2}y/dx^{2} = -0.00143[/tex]

[tex]x = 1023.785ft[/tex]

[tex]\rho = \frac{1 + ((\frac{dy}{dx})^{2})^{3/2}}{|d^{2}y/dx^{2}|}[/tex]

[tex]\rho = \frac{1 + (((-0.00143)(1023.785 ft))^{2})^{3/2}}{|-0.00143|}[/tex]

[tex]\rho = 3900.339ft[/tex] I am assuming this is the [tex]\rho[/tex] for (b).

I am unsure how to get the [tex]\rho[/tex] for (a).

The [tex]a_{t}[/tex] I think for (a) is 0 because the velocity in the horizontal is constant. I am unsure the [tex]a_{n}[/tex] because I need [tex]\rho[/tex].

I don't understand how to get [tex]a_{t}[/tex] and [tex]a_{n}[/tex] just before it strikes the ground because I don't have a time, do I just assume 1 second before hitting the ground?

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