A circus performer stands on a platform and throws an apple from a height of 33.0 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 = 23.0 m/s,
 how far from the end of the platform should the second performer stand? (Assume 𝜃 = 20.0°.)

To determine how far the second performer should stand, we need to find the **horizontal distance** the apple travels before it hits the ground. This involves breaking down the problem using the principles of **projectile motion**.

### Key Information:
- Initial height: \( h_0 = 33.0 \, \text{m} \)
- Initial velocity: \( v_0 = 23.0 \, \text{m/s} \)
- Launch angle: \( \theta = 20.0^\circ \)
- Gravity: \( g = 9.81 \, \text{m/s}^2 \)

### Step 1: Decompose the initial velocity
We'll break the initial velocity into horizontal and vertical components:
- \( v_{0x} = v_0 \cos(\theta) \) (horizontal component)
- \( v_{0y} = v_0 \sin(\theta) \) (vertical component)

\[
v_{0x} = 23.0 \, \text{m/s} \times \cos(20.0^\circ) 
\]
\[
v_{0y} = 23.0 \, \text{m/s} \times \sin(20.0^\circ) 
\]

### Step 2: Time of flight
We can calculate the time the apple spends in the air by solving for the time \( t \) it takes for the apple to fall from 33.0 m, considering its initial upward velocity and downward acceleration due to gravity.

Using the vertical displacement formula:

\[
y(t) = v_{0y} t - \frac{1}{2} g t^2 + h_0
\]

Where \( y(t) = 0 \) (since it hits the ground), we solve for \( t \).

### Step 3: Horizontal distance
Once we know the time of flight \( t \), the horizontal distance the apple travels is:

\[
x = v_{0x} \cdot t
\]

Let's calculate these steps now!It seems there's an issue with the calculation as the horizontal distance turned out to be negative. Let's correct the equation setup and reattempt.

In the quadratic equation for the time of flight, we should ensure the correct signs for gravity and vertical motion terms. I'll correct and calculate again.The horizontal distance the apple travels is approximately **76.01 meters**. Therefore, the second performer should stand about **76 meters** away from the end of the platform to catch the apple.

Would you like further details on any part of this solution, or do you have additional questions?

Here are some related questions:
1. How does air resistance affect projectile motion?
2. How can we determine the maximum height reached by the apple?
3. What if the performer throws the apple at a different angle? How would the distance change?
4. What happens if the initial height changes?
5. How does changing gravity (e.g., on another planet) affect the horizontal distance?

**Tip:** In projectile motion, the horizontal and vertical motions are independent of each other, with the only link being the time of flight.

A circus performer stands on a platform and throws an apple from a height of 33.0 m above the ground with an initial velocity v0 = 23.0 m/s, how far from the end of the platform should the second performer stand? (Assume 𝜃 = 20.0°.)

Solve a projectile motion problem involving an apple thrown from a 33.0 m high platform with an initial velocity of 23.0 m/s at an angle of 20°. Learn how to break the motion into horizontal and vertical components, calculate the time of flight, and determine the horizontal distance. Suitable for Grades 10-12 physics students.

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