A circus performer stands on a platform and throws an apple from a height of 40.0 m above the ground with an initial velocity v0 as shown in the figure below. A second, blindfolded performer must catch the apple. If v0 = 17.0 m/s, how far from the end of the platform should the second performer stand? (Assume 𝜃 = 30.0°.) use v0y=-v0sinθ since the apple is thrown at a downward angle

Let's break down the problem to determine how far the second performer should stand to catch the apple.

### Given Information:
- Height of the platform: \( h = 40.0 \, \text{m} \)
- Initial velocity: \( v_0 = 17.0 \, \text{m/s} \)
- Angle of projection: \( \theta = 30.0^\circ \) (measured downward from the horizontal)
- Acceleration due to gravity: \( g = 9.81 \, \text{m/s}^2 \)

We need to find how far from the platform (horizontally) the second performer should stand, denoted as \( x \).

### Step 1: Break the initial velocity into components
We decompose the initial velocity \( v_0 \) into horizontal and vertical components:

- **Horizontal component**:
  \[
  v_{0x} = v_0 \cos(\theta)
  \]
  \[
  v_{0x} = 17.0 \times \cos(30^\circ) = 17.0 \times 0.866 = 14.72 \, \text{m/s}
  \]

- **Vertical component**:
  Since the apple is thrown downward, the vertical component is:
  \[
  v_{0y} = -v_0 \sin(\theta)
  \]
  \[
  v_{0y} = -17.0 \times \sin(30^\circ) = -17.0 \times 0.5 = -8.5 \, \text{m/s}
  \]

### Step 2: Time of flight
We use the vertical motion equation to determine the time the apple is in the air:
\[
y = y_0 + v_{0y} t + \frac{1}{2} g t^2
\]
Where:
- \( y_0 = 40.0 \, \text{m} \)
- \( y = 0 \) (ground level)
- \( v_{0y} = -8.5 \, \text{m/s} \)
- \( g = 9.81 \, \text{m/s}^2 \)

Substitute values:
\[
0 = 40.0 - 8.5 t + \frac{1}{2} \times 9.81 \, t^2
\]
Simplify:
\[
4.905 t^2 - 8.5 t - 40 = 0
\]
This is a quadratic equation. Solving it using the quadratic formula:
\[
t = \frac{-(-8.5) \pm \sqrt{(-8.5)^2 - 4 \times 4.905 \times (-40)}}{2 \times 4.905}
\]
\[
t = \frac{8.5 \pm \sqrt{72.25 + 784.8}}{9.81}
\]
\[
t = \frac{8.5 \pm \sqrt{857.05}}{9.81}
\]
\[
t = \frac{8.5 \pm 29.28}{9.81}
\]
We take the positive root since time must be positive:
\[
t = \frac{8.5 + 29.28}{9.81} = \frac{37.78}{9.81} \approx 3.85 \, \text{seconds}
\]

### Step 3: Horizontal distance traveled
Now that we have the time, the horizontal distance \( x \) can be calculated using:
\[
x = v_{0x} t
\]
\[
x = 14.72 \times 3.85 \approx 56.7 \, \text{m}
\]

### Final Answer:
The second performer should stand **56.7 meters** from the end of the platform to catch the apple.

Would you like more details or have any questions?

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Here are five related questions you might find useful:
1. What happens if the apple is thrown at a higher angle?
2. How would the horizontal distance change if the initial speed increased?
3. What if the second performer stands on an elevated platform? How would that affect the calculation?
4. What is the role of air resistance in this calculation?
5. How would the result change if the apple was thrown at an upward angle?

**Tip:** When dealing with projectiles, always decompose the velocity into horizontal and vertical components, as it simplifies calculations involving motion.

Learn how to solve a projectile motion problem where an apple is thrown from a 40 m high platform with an initial velocity of 17.0 m/s at a 30° downward angle. This problem covers key concepts like decomposing velocity into components, solving quadratic equations, and calculating the horizontal distance the apple travels. Suitable for Grades 10-12.

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