A package is dropped from an airplane flying horizontally with constant speed V in the positive xdirection. The package is released at time t = 0 from a height H above the origin. In addition to the vertical component of acceleration due to gravity, a strong wind blowing from the right gives the package a horizontal component of acceleration of magnitude ¼g to the left. Derive an expression for the horizontal distance D from the origin where the package hits the ground.

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lcmendozaf lcmendozaf · Answer 1 · 2019-09-14T15:42:34+02:00

Answer:

[tex]D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}[/tex]

Explanation:

From the vertical movement, we know that initial speed is 0, and initial height is H, so:

[tex]Y_{f}=Y_{o}-g*\frac{t^{2}}{2}[/tex]

[tex]0=H-g*\frac{t^{2}}{2}[/tex] solving for t:

[tex]t=\sqrt{\frac{2H}{g} }[/tex]

Now, from the horizontal movement, we know that initial speed is V and the acceleration is -g/4:

[tex]X_{f}=X_{o}+V*t+a*\frac{t^{2}}{2}[/tex] Replacing values:

[tex]D=V*\sqrt{\frac{2H}{g} }-\frac{g}{4}*\frac{1}{2} *(\sqrt{\frac{2H}{g} })^{2}[/tex]

Simplifying:

[tex]D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}[/tex]