Question

A 415-kg container of food and water is dropped from an airplane at an altitude of 300m. First, consider the situation ignoring air resistance. Then calculate the more realistic situation involving a drag force provided by a parachute.
1. If you ignore air resistance, how long will it take the container to fall 300m to the ground?

2. Again, ignoring air resistance, what is the speed of the container just before it hits the ground?

3. The container is attached to a parachute designed to produce a drag force that allows the container to reach a constant downward velocity of 6m/s. What is the magnitude of the drag force when the container is falling at a constant 6m/s down?

Answer 1

1) 7.8 s

Ignoring air resistance, the only force acting on the container is gravity. Therefore, we can use the following suvat equation for the free-fall case:

$s=ut+\frac{1}{2}at^2$

where

s = 300 m is the displacement of the container

u = 0 is the initial vertical velocity of the container, which starts from rest

a = g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Here we have chosen downward as positive direction, so all quantities are positive

Solving the equation for t, we find:

$s=\frac{1}{2}gt^2\\t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(300)}{9.8}}=7.8 s$

2) 76.4 m/s

The speed of the container at any time t can be found by using another suvat equation:

$v=u+at$

where

v is the speed at time t

u = 0 is the initial speed

a = g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Substituting t = 7.8 s, the time of flight, we can immediately find the final speed of the container, just before hitting the ground:

$v=0+(9.8)(7.8)=76.4 m/s$

3) 4067 N

In this case, there is a parachute that produces a drag force acting upward.

When the container is falling at constant speed, 6 m/s down, it means that its acceleration is zero, so the net force acting on it is zero. This means that the air drag balances the weight of the container, therefore we can write:

$F=mg$

where

F is the air drag

m is the mass of the container

g is the acceleration of gravity

Substituting m = 415 kg, we find the magnitude of the drag force:

$F=(415)(9.8)=4067 N$