Question

A hot-air balloon is rising upward with a constant speed of 2.95 m/s. When the balloon is 3.30 m above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?

Answer 1

Consider the compass: you know the initial vertical velocity, its initial height, and gravitational acceleration.

hfinal=hinitial+ vi*time -4.9 t^2
solve for t. hfinal is 0, hinitial is 3.30

Answer 2

5m/s

Answer 3

[AI-generated answer — Explain Bot]

To find the time it takes for the compass to hit the ground, we can use the equation of motion for vertical motion:

𝑑 = 𝑣𝑖𝑛𝑖𝑡 + 0.5𝑎𝑡²

where:
𝑑 = distance travelled (in this case, 3.30 m)
𝑣𝑖 = initial velocity (in this case, 0 m/s since the compass is dropped)
𝑎 = acceleration (in this case, the acceleration due to gravity, which is approximately 9.8 m/s²)
𝑡 = time (what we're trying to find)

Rearranging the equation, we get:

0.5𝑎𝑡² + 𝑣𝑖𝑛𝑖𝑡 - 𝑑 = 0

Now we can substitute the known values into the equation:

0.5(9.8 𝑡)² + (0)(𝑡) - 3.30 = 0

Simplifying the equation, we have:

4.9𝑡² - 3.30 = 0

Now, we can solve this quadratic equation using the quadratic formula:

𝑡 = (-𝑏 ± √(𝑏² - 4𝑎𝑐)) / (2𝑎)

In this case, 𝑎 = 4.9, 𝑏 = 0, and 𝑐 = -3.30. Plugging these values into the formula, we get:

𝑡 = ± √((0)² - 4(4.9)(-3.30)) / (2(4.9))

Simplifying further, we have:

𝑡 = ± √(0 + 64.68) / 9.8

𝑡 = ± √(64.68) / 9.8

Since time cannot be negative, we take the positive square root:

𝑡 = √(64.68) / 9.8

Evaluating this using a calculator, we find:

𝑡 ≈ 1.20 s

Therefore, it takes approximately 1.20 seconds for the compass to hit the ground.