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 = 15.0 m/s,
 how far from the end of the platform should the second performer stand? (Assume 𝜃 = 30.0°.)

We are dealing with a projectile motion problem where a circus performer throws an apple with an initial velocity of \(v_0 = 15.0 \, \text{m/s}\) at an angle \(\theta = 30^\circ\) from a platform at a height \(h = 33.0 \, \text{m}\). Our goal is to determine how far the second performer should stand from the end of the platform to catch the apple.

### Step 1: Split the initial velocity into components
We first break down the initial velocity into its horizontal (\(v_{0x}\)) and vertical (\(v_{0y}\)) components using trigonometry:

\[
v_{0x} = v_0 \cos \theta
\]
\[
v_{0y} = v_0 \sin \theta
\]

Given \(v_0 = 15.0 \, \text{m/s}\) and \(\theta = 30^\circ\):

\[
v_{0x} = 15.0 \cos 30^\circ = 15.0 \times \frac{\sqrt{3}}{2} \approx 12.99 \, \text{m/s}
\]
\[
v_{0y} = 15.0 \sin 30^\circ = 15.0 \times \frac{1}{2} = 7.5 \, \text{m/s}
\]

### Step 2: Time of flight
Now, we calculate the total time the apple is in the air. The vertical motion follows this equation:

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

where \(y = -33.0 \, \text{m}\) (since the apple falls 33 meters below the platform), \(v_{0y} = 7.5 \, \text{m/s}\), and \(g = 9.8 \, \text{m/s}^2\) is the acceleration due to gravity.

We solve the quadratic equation for \(t\). The equation becomes:

\[
-33 = 7.5 t - \frac{1}{2} \times 9.8 \times t^2
\]

Simplifying:

\[
-33 = 7.5 t - 4.9 t^2
\]

Rearranging it into a standard quadratic form:

\[
4.9 t^2 - 7.5 t - 33 = 0
\]

We solve this quadratic equation for \(t\).The time of flight \( t \) is approximately \( 3.47 \, \text{seconds} \).

### Step 3: Horizontal distance
Now that we know how long the apple is in the air, we can calculate the horizontal distance it travels. The horizontal velocity component \( v_{0x} \) remains constant because there is no horizontal acceleration.

The horizontal distance \( d \) is given by:

\[
d = v_{0x} \times t
\]

Substituting \( v_{0x} = 12.99 \, \text{m/s} \) and \( t = 3.47 \, \text{seconds} \):The second performer should stand approximately \( 45.1 \, \text{meters} \) from the end of the platform to catch the apple.

Would you like any further details or clarifications?

Here are five related questions you might find useful:
1. What happens if the initial velocity of the apple increases?
2. How does changing the launch angle affect the horizontal distance?
3. What if the platform height is different, how would that impact the outcome?
4. How would air resistance alter the calculation?
5. Can the apple still reach the performer if they stand closer than the calculated distance?

**Tip:** Always break down projectile motion into horizontal and vertical components for easier analysis.

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 = 15.0 m/s. How far from the end of the platform should the second performer stand? (Assume 𝜃 = 30.0°.)

Solve a detailed projectile motion problem where a circus performer throws an apple with an initial velocity of 15.0 m/s at a 30° angle from a platform 33.0 m high. Learn how to calculate the horizontal distance for the second performer to stand using trigonometry, kinematics, and quadratic equations. Suitable for Grades 10-12.

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