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:
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:
2) 76.4 m/s
The speed of the container at any time t can be found by using another suvat equation:
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:
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:
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: