Question

An airplane is flying with a velocity of v0 at an angle of α above the horizontal. When the plane is a distance h directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment.

Part A
How far from the dog will the suitcase land? You can ignore air resistance.
Take the free fall acceleration to be g.

Answer 1

Answer:

$s = v_0\sqrt{\frac{2\alpha}{g}}$

Explanation:

When the suitcase is being released, it has a horizontal velocity of v0, a vertical velocity of 0 and vertical acceleration of g

The time it takes for it to drop by the vertical distance of α under acceleration g can be solved using the following equation of motion

$\alpha = gt^2/2$

$t^2 = \frac{2\alpha}{g}$

$t = \sqrt{\frac{2\alpha}{g}}$

This is also the time it takes for the suitcase to travel horizontally at the rate of v0. So the horizontal distance it travels is

$s = v_0t = v_0\sqrt{\frac{2\alpha}{g}}$