At a particular instant, a hot air balloon is 100 m in the air and descending at a constant speed of 2.0 m/s. At this exact instant, a girl throws a ball horizontally, relative to herself, with an initial speed of 16 m/s. When she lands, where will she find the ball? Ignore air resistance.

To solve this problem, we need to analyze both the horizontal and vertical motions separately, as the ball is thrown horizontally and air resistance is ignored. 

### Step 1: Analyze vertical motion
Since the hot air balloon is descending at a constant speed of 2.0 m/s, the ball has an initial vertical velocity of \( -2.0 \, \text{m/s} \) (downward) at the moment it's released.

The vertical motion follows the equations of motion under constant acceleration (gravity):

\[
y(t) = y_0 + v_{y0} t + \frac{1}{2} a t^2
\]
Where:
- \( y_0 = 100 \, \text{m} \) (initial height of the ball),
- \( v_{y0} = -2.0 \, \text{m/s} \) (initial vertical velocity),
- \( a = -9.8 \, \text{m/s}^2 \) (acceleration due to gravity).

The ball will hit the ground when \( y(t) = 0 \).

Setting up the equation:
\[
0 = 100 - 2.0 t - 4.9 t^2
\]
This is a quadratic equation. Rearranging it:
\[
4.9 t^2 + 2.0 t - 100 = 0
\]

### Step 2: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where \( a = 4.9 \), \( b = 2.0 \), and \( c = -100 \).

Let me calculate that.The two solutions for time are:

\[
t_1 \approx 4.32 \, \text{seconds} \quad \text{(positive, valid solution for time)}
\]
\[
t_2 \approx -4.73 \, \text{seconds} \quad \text{(negative, not physically meaningful)}
\]

Therefore, the time it takes for the ball to hit the ground is approximately \( t = 4.32 \, \text{seconds} \).

### Step 3: Analyze horizontal motion
Since the ball is thrown horizontally with an initial speed of \( 16 \, \text{m/s} \) and there is no air resistance, the horizontal velocity remains constant. The horizontal distance the ball travels is given by:

\[
x = v_x \cdot t
\]

Where:
- \( v_x = 16 \, \text{m/s} \) (horizontal speed),
- \( t = 4.32 \, \text{seconds} \).

Now, let's calculate the horizontal distance.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can calculate the horizontal distance as follows:

\[
x = v_x \cdot t = 16 \, \text{m/s} \times 4.32 \, \text{seconds} \approx 69.12 \, \text{meters}
\]

### Final answer:
The girl will find the ball approximately 69.12 meters horizontally away from where she was when she threw it.

If you need further details or clarifications, feel free to ask!

Here are some related questions:
1. What if the balloon were rising instead of descending?
2. How would air resistance affect the trajectory?
3. How does the initial speed of the ball affect where it lands?
4. Can you calculate the time for different initial vertical velocities?
5. What happens if the girl throws the ball at an angle instead of horizontally?

**Tip:** Breaking a complex motion into horizontal and vertical components makes it easier to solve projectile problems.

Solve a real-world projectile motion problem where a ball is thrown horizontally from a descending hot air balloon. Learn how to calculate the time of flight using quadratic equations and how far the ball will land horizontally. This problem involves kinematics, suitable for students in Grades 10-12.

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