A hot-air balloon is rising upward with a constant speed of 2.50 m/s. When the balloon is 3.00 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 ?

Options

A. 2.43 s
B. 0.568 s
C. 1.08 s
D. 0.410 s

Answer

To solve the problem of the hot-air balloon and the dropped compass, we can follow these steps:

### Step 1: Understand the initial conditions
The hot-air balloon is rising with a constant speed of \( u = 2.50 \, \text{m/s} \) when the compass is dropped from a height of \( h = 3.00 \, \text{m} \) above the ground.

### Step 2: Determine the initial velocity of the compass
When the compass is dropped, it has the same initial velocity as the balloon, which is \( u = -2.50 \, \text{m/s} \) (the negative sign indicates that it is moving upwards, opposite to the downward direction we will consider as positive).

### Step 3: Set up the equation of motion
We will use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) is the displacement (3 m, downward, so we will take it as positive),
- \( u \) is the initial velocity of the compass (-2.50 m/s),
- \( a \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \), taken as positive since we are considering downward as positive),
- \( t \) is the time in seconds.

Substituting the known values into the equation:
\[ 3 = (-2.50)t + \frac{1}{2}(9.81)t^2 \]

### Step 4: Rearrange the equation
Rearranging gives us:
\[ 0 = \frac{1}{2}(9.81)t^2 - 2.50t - 3 \]
Multiplying through by 2 to eliminate the fraction:
\[ 0 = 9.81t^2 - 5.00t - 6 \]

### Step 5: Solve the quadratic equation
We can apply the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 9.81 \), \( b = -5.00 \), and \( c = -6 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-5.00)^2 - 4(9.81)(-6) = 25 + 235.44 = 260.44 \]

Now substituting into the quadratic formula:
\[ t = \frac{5.00 \pm \sqrt{260.44}}{2 \times 9.81} \]
Calculating \( \sqrt{260.44} \approx 16.13 \):
\[ t = \frac{5.00 \pm 16.13}{19.62} \]

### Step 6: Calculate the possible values for \( t \)
Calculating the two possible values:
1. \( t = \frac{5.00 + 16.13}{19.62} \approx \frac{21.13}{19.62} \approx 1.08 \, \text{s} \)
2. \( t = \frac{5.00 - 16.13}{19.62} \) gives a negative value, which we discard since time cannot be negative.

### Final Answer
Thus, the time elapsed before the compass hits the ground is approximately:
\[ \boxed{1.08 \, \text{s}} \]