Question

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.

Answer 1

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:

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