An airplane is flying at a constant height of 3000 ft abovewater at a speed of 400 ft/s

Chapter 0, Problem 55

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An airplane is flying at a constant height of \(3000ft\)above water at a speed of \(400ft/s\). The pilot is to release a survival package so that it lands in the water at a sighted point \(P\). If air resistance is neglected, then the package will follow a parabolic trajectory whose equation relative to the coordinate system in the accompanying figure is

                                    \(y=3000-\frac{g}{2 v^{2}} x^{2}\)

where \(g\) is the acceleration due to gravity and \(v\) is the speed of the airplane. Using \(g=32 \mathrm{ft} / \mathrm{s}^{2}\), find the "line of sight" angle \(\theta\), to the nearest degree, that will result in the package hitting the target point.

Equation Transcription:

Text Transcription:

3000ft

400ft/s

P

y=3000-g/2v^2 x^2

g

v

g=32ft/s^2