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nasty-shy [4]
4 years ago
12

A package is dropped from an airplane flying horizontally with constant speed V in the positive xdirection. The package is relea

sed 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.
Physics
1 answer:
k0ka [10]4 years ago
4 0

Answer:

D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}

Explanation:

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

Y_{f}=Y_{o}-g*\frac{t^{2}}{2}

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

t=\sqrt{\frac{2H}{g} }

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

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

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

Simplifying:

D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}

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Explanation:

Because the spaceships are traveling in the same direction, you should subtract the speed of spaceship A from the speed of spaceship B in order to calculate v_{AB}.

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