Question

A practical problem! Your friend shipped out on a cruise aboard the Coral Sea and forgot their Gameboy (that they need to test dolphin cognition). You and your drone pilot friend quickly devise a plan to airdrop the vital instrument package before the ship gets too far offshore. With good fortune deploy your plan on a windless day. You devise a spherical delivery package of closed cell foam to protect the instrument from impact, and float (you hope) if it does not land on the ship. The package weighs 6 kg and has the same diameter as a basketball. a. Safety regulations prohibit drones from flying closer than 200 ft (61 m) above the water surface near ships. The Coral Sea’s main deck sits 2 m above water level. Calculate how long it would for your package to land on deck after dropped from the drone at 61 m above water level. b.Help your precision pilot: Your drone catches up with the Coral Sea to find it cruising at 10 knots (5.1 m/s). If your pilot matches the speed of the Coral Sea, then as soon as you drop the package, it experiences a drag force in the horizontal direction from the 5.1 m/s airflow that the drone had a moment ago pulled it through. This will of course slow its horizontal movement from this drag force. Remember that F=ma so that you can calculate how quickly it slows down. You remember that if you know an initial velocity, an object’s acceleration, and how long it accelerates that you can calculate the final velocity for that time. Use your time in free fall that you calculated above to estimate how much the package slows from it’s initial cruising speed during free fall to estimate how far ahead of the ship that your friend should pilot the drone to compensate for this change in horizontal speed to get your package to land on board the Coral Sea when dropped? Show full worked out solutions and calculations, including all intermediary steps (IE drag forces, velocities etc)

Answer 1

height of fall = 61 - 2 = 59m

using second kinematical equation

s = ut + 1/2 at^2

59 = 0 + 1/2 x 9.8 x t^2

t = 3.47 sec ... time taken to cover the vertical displacement

The drag force on a spherical object is given by

Fd =

= 1.85 x 10^-5 ... viscosity of air

r = 12 cm = 0.12m ... radius of basketball

v = 5.1 m/s ... velocity

Fd = 6 x 3.14 x 1.85 x 10^-5 x 0.12 x 5.1

= 2.13 x 10^-4 N

decelereation = F/m = 2.13 x 10^-4 / 6 = 3.55 x 10^-5 m/s^2

using second kinematical equation

s = ut + 1/2 at^2

= 0 - 1/2 x 3.55 x 10^-5 x 3.47^2 .... u = 0 because intitially the drone and cruise were moving at same speed. Hence their relative speed were 0

= -2.14 x 10^-4 m

This is the distance that packet will fall behind, hence the drone must be horizontally this distance ahead of the cruise.