A hot air balloon just lifted off and is rising at a constant rate of 2m/s. A friend picks it up and tosses it straight upward with an initial speed of 12 m/s. If the passenger is 2.5m above her friend when the camera is tossed, how high is she above her friend when the camera reaches her?

To solve this problem, we need to consider the motion of both the hot air balloon (with the passenger) and the camera separately and then find the point at which they meet.

### Step 1: Motion of the Hot Air Balloon
- The balloon is rising at a constant rate of \( v_b = 2 \, \text{m/s} \).
- The initial height of the passenger above her friend when the camera is tossed is \( h_0 = 2.5 \, \text{m} \).
- The height of the passenger after \( t \) seconds is given by:
\[
h_b(t) = h_0 + v_b \cdot t = 2.5 + 2t \, \text{m}
\]

### Step 2: Motion of the Camera
- The camera is tossed with an initial upward speed of \( v_c = 12 \, \text{m/s} \).
- The initial height of the camera is \( h_c(0) = 0 \, \text{m} \) (it's tossed from ground level).
- The height of the camera after \( t \) seconds is given by the equation for uniformly accelerated motion:
\[
h_c(t) = v_c \cdot t - \frac{1}{2} g t^2
\]
where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

Substituting the known values:
\[
h_c(t) = 12t - \frac{1}{2} \cdot 9.8 \cdot t^2 = 12t - 4.9t^2 \, \text{m}
\]

### Step 3: Equating Heights to Find the Time
The camera reaches the passenger when the heights of the balloon and the camera are equal:
\[
h_b(t) = h_c(t)
\]
Substituting the expressions we derived:
\[
2.5 + 2t = 12t - 4.9t^2
\]
Rearranging this into a standard quadratic equation:
\[
4.9t^2 - 10t + 2.5 = 0
\]

### Step 4: Solving the Quadratic Equation
We can solve this quadratic equation using the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 4.9 \), \( b = -10 \), and \( c = 2.5 \).

Let's calculate \( t \):

\[
t = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 4.9 \cdot 2.5}}{2 \cdot 4.9}
\]
\[
t = \frac{10 \pm \sqrt{100 - 49}}{9.8}
\]
\[
t = \frac{10 \pm \sqrt{51}}{9.8}
\]

Calculating \( \sqrt{51} \approx 7.14 \):

\[
t = \frac{10 \pm 7.14}{9.8}
\]

This gives two potential solutions:

\[
t_1 = \frac{17.14}{9.8} \approx 1.75 \, \text{seconds}, \quad t_2 = \frac{2.86}{9.8} \approx 0.29 \, \text{seconds}
\]

Since we're looking for when the camera reaches the passenger, we take the positive value \( t_1 = 1.75 \) seconds.

### Step 5: Finding the Height
Now, substitute \( t = 1.75 \) seconds back into the equation for the balloon's height:

\[
h_b(1.75) = 2.5 + 2 \cdot 1.75 = 2.5 + 3.5 = 6 \, \text{meters}
\]

### Conclusion:
The passenger is **6 meters** above her friend when the camera reaches her.

Would you like further details or have any questions?

Here are 5 related questions:
1. How would the problem change if the initial speed of the camera was different?
2. What happens if the balloon is also accelerating upwards?
3. How does air resistance affect the camera's trajectory?
4. What if the camera is thrown at an angle instead of straight up?
5. How would the time differ if the balloon started at a higher initial height?

**Tip:** When solving problems involving multiple moving objects, it's often useful to write equations of motion for each object separately before combining them to find when and where they meet.

Explore a physics problem involving a hot air balloon and a camera, where the camera is tossed upward while the balloon rises at a constant rate. Learn how to calculate the height of the passenger when the camera reaches her, using equations of motion and quadratic equations.

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