a hot air balloonist, rising vertically with a constant velocity of magnitude 5m/s releases a sandbag at an instant when the balloon is 40m above the ground. after it is released, the sandbag is in free fall.

Question

a) compute position at .250s
b.)how many seconds after its release, will it strike the ground?
c.)is free fall related to inertia that much that i should consider the newtonian law when answering this question?

Answer 1

The height at time t after release is, in meters,

H = 40 + 5 t - (g/2)t^2
g = 9.8 m/s^2 in this case

(a) Plug t = 0.25 s into that formula
(b) Solve 40 + 5t - 4.9 t^2 = 0 to get the t when it his the ground.
(c) Yes, you must consider the Newtonian Laws of motion whether in free fall or not.

Answer 2

why is 5m/s used as initial velocity of sandbag? if inertia was to be considered, what would be the sandbags initial volume then?
i used the formula:

Yf-Yi=Vit + [(-gt^2)/2]

my Vi was 0m/s since i assumed that the sandbag was at rest.

Answer 3

When the sandbag is released, it has the 5 m/s upward velocity of the ballon that is carrying it.

The volume of the sandbag is not asked for, and has nothing to do with the problem. Perhaps you meant Velocity

Answer 4

How did the sandbag gain a velocity of 5m/s upward? I thought of the sandbag remained at rest?

Answer 5

i meant of velocity. let me reword that: if inertia was to be considered, what would be the sandbags initial velocity then?

thanks for enlightening me.

Answer 6

Inertia IS considered. That is why the sandbag retains an initial vertical velocity of 5 m/s when it is dumped overboard. Then it starts accelerating downward because of gravity. Before that, it was being pushed upward with a force equal to the weight, but whatever rope or shelf was holdng it to the ballon