Question

 A rescue pilot drops a survival kit while her plane is flying at an altitude of 1200.0 meters with a forward velocity of 97.4 m/s. How far in advance of the drop zone should she release the package?

Answer 1

Time of falling $t$ can be found from the kinematic equation (if there is no vertical initial velocity component):

$$h = \dfrac{gt^2}{2}$$

where $h = 1200m$ is the altitude, and $g = 9.81m/s^2$ is the gravitational acceleration. Thus, obtain:

$$t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2\cdot 1200}{9.81}} \approx 15.6s$$

In this time the survival kit covers the following horizontal distance:

$$d = v_0t = 97.4m/s\cdot 15.6s \approx 1523m$$

Here $v_0 = 97.4m/s$ is the initial horizontal velocity of the kit. Thus, it should be release 1523 m in advance.

Answer. 1523 m.