Question

A balloonist, riding in the basket of a hot air balloon that is rising vertically with a constant velocity of 10.1 m/s, releases a sandbag when the balloon is 43.4 m above the ground. Neglecting air resistance, what is the bag's speed when it hits the ground? Assume g = 9.80 m/s2.

m/s

Answer 1

Larry, or Jake: I will be happy to critique your thinking.

Answer 2

Tried using a free fall set up with no luck. Things like (-9.8)(10.1)+43.3

Answer 3

You can get a quick answer using energy conservation.

(1/2)M V2^2 = M g H + (1/2) M V1^1

M's cancel.

V2 = sandbag when it hits the ground
V1 = sandbag velocity when released (relative to ground) = 10.1 m/s
H = 43.4 m

Answer 4

30.87 m/s

Answer 5

answwr

Answer 6

[AI-generated answer — Explain Bot]

To find the speed of the sandbag when it hits the ground, we can use the equations of motion.

1. First, let's determine the time it takes for the sandbag to fall from a height of 43.4 m. We can use the equation:

h = (1/2)gt^2

where h is the height, g is the acceleration due to gravity, and t is the time. Rearranging this equation, we get:

t^2 = (2h) / g

Substituting the given values, we have:

t^2 = (2 * 43.4) / 9.80

t^2 = 8.8367

Taking the square root of both sides, we find:

t ≈ 2.97 seconds

2. Now, we can calculate the final velocity of the sandbag using the equation:

v = gt

Substituting the values, we have:

v = 9.80 * 2.97

v ≈ 29.06 m/s

Therefore, the speed of the sandbag when it hits the ground is approximately 29.06 m/s.