Question

A military plane wants to drop supplies to some survivors on a rocky ndge. plane is flying horizontally at an altitude of 235 m with a speed of 72.5 m/s . How far in advance must the relief goods be released so that they arrive precisely at the survivors' location?

Answer 1

The initial speed in horizontal direction is $v_{0h} = 72.5m/s$. This speed does not change with time. The initial speed in vertical direction is $v_{0v} = 0m/s$. Thus, the distance travelled in vertial direction is

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

where if $h = 235m$ then $t$ is the time of falling. Thus:

$$t = \sqrt{\dfrac{2h}{g}}$$

Then the distance travelled in horizontal direction is:

$$d = v_{0h}t = v_{0h}\sqrt{\dfrac{2h}{g}}\\ d = 72.5\sqrt{\dfrac{2\cdot 235}{9.81}} \approx 501.8 m$$

Answer. The relief goods must be released 501.8 m in advance.