Math Problem Statement

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 πœƒ = 30.0Β°.)

Solution

We are dealing with a projectile motion problem where a circus performer throws an apple with an initial velocity of v0=15.0 m/sv_0 = 15.0 \, \text{m/s} at an angle ΞΈ=30∘\theta = 30^\circ from a platform at a height h=33.0 mh = 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 (v0xv_{0x}) and vertical (v0yv_{0y}) components using trigonometry:

v0x=v0cos⁑θv_{0x} = v_0 \cos \theta v0y=v0sin⁑θv_{0y} = v_0 \sin \theta

Given v0=15.0 m/sv_0 = 15.0 \, \text{m/s} and ΞΈ=30∘\theta = 30^\circ:

v0x=15.0cos⁑30∘=15.0Γ—32β‰ˆ12.99 m/sv_{0x} = 15.0 \cos 30^\circ = 15.0 \times \frac{\sqrt{3}}{2} \approx 12.99 \, \text{m/s} v0y=15.0sin⁑30∘=15.0Γ—12=7.5 m/sv_{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=v0ytβˆ’12gt2y = v_{0y} t - \frac{1}{2} g t^2

where y=βˆ’33.0 my = -33.0 \, \text{m} (since the apple falls 33 meters below the platform), v0y=7.5 m/sv_{0y} = 7.5 \, \text{m/s}, and g=9.8 m/s2g = 9.8 \, \text{m/s}^2 is the acceleration due to gravity.

We solve the quadratic equation for tt. The equation becomes:

βˆ’33=7.5tβˆ’12Γ—9.8Γ—t2-33 = 7.5 t - \frac{1}{2} \times 9.8 \times t^2

Simplifying:

βˆ’33=7.5tβˆ’4.9t2-33 = 7.5 t - 4.9 t^2

Rearranging it into a standard quadratic form:

4.9t2βˆ’7.5tβˆ’33=04.9 t^2 - 7.5 t - 33 = 0

We solve this quadratic equation for tt.The time of flight tt is approximately 3.47 seconds3.47 \, \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 v0xv_{0x} remains constant because there is no horizontal acceleration.

The horizontal distance dd is given by:

d=v0xΓ—td = v_{0x} \times t

Substituting v0x=12.99 m/sv_{0x} = 12.99 \, \text{m/s} and t=3.47 secondst = 3.47 \, \text{seconds}:The second performer should stand approximately 45.1 meters45.1 \, \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.

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Math Problem Analysis

Mathematical Concepts

Projectile Motion
Trigonometry
Quadratic Equations
Kinematics

Formulas

v_{0x} = v_0 \cos \theta
v_{0y} = v_0 \sin \theta
y = v_{0y} t - \frac{1}{2} g t^2
d = v_{0x} \times t
Quadratic equation: ax^2 + bx + c = 0

Theorems

Kinematic Equations of Motion
Quadratic Formula

Suitable Grade Level

Grades 10-12