Math Problem Statement

A 75kg long jumper uses two weights of 5.1kg, throwing them horizontally to the rear at his maximum height such that their horizontal velocity is zero relative to the ground. Let his liftoff velocity be 9.3i+4.1j m/s with or without the weights, and assume that he lands at the liftoff level. What distance would the weights add to his range. the answer is not 108cm

Solution

To determine the additional distance the weights contribute to the long jumper’s range, we apply the principle of conservation of momentum and projectile motion.

Step 1: Define the Given Values

  • Mass of the jumper + weights: M=75+2(5.1)=85.2M = 75 + 2(5.1) = 85.2 kg
  • Mass of the jumper after throwing weights: mj=75m_j = 75 kg
  • Mass of each weight: mw=5.1m_w = 5.1 kg
  • Initial velocity: vi=(9.3i^+4.1j^)\mathbf{v}_i = (9.3 \hat{i} + 4.1 \hat{j}) m/s
  • Final velocity of weights: vw=0\mathbf{v}_w = 0 (since they stop moving relative to the ground)

Step 2: Apply Conservation of Momentum (Horizontal)

Since no external horizontal forces act on the system, the total momentum before and after throwing the weights must remain the same.

Before Throwing the Weights:

Total horizontal momentum:

Pinitial=MvixP_{\text{initial}} = M v_{ix}

Pinitial=(85.2)(9.3)P_{\text{initial}} = (85.2)(9.3)

Pinitial=792.36 kg\cdotpm/sP_{\text{initial}} = 792.36 \text{ kg·m/s}

After Throwing the Weights:

The weights come to rest relative to the ground, so they contribute no horizontal momentum. The jumper’s new horizontal velocity vfxv_{fx} satisfies:

mjvfx=Pinitialm_j v_{fx} = P_{\text{initial}}

75vfx=792.3675 v_{fx} = 792.36

vfx=792.3675=10.565 m/sv_{fx} = \frac{792.36}{75} = 10.565 \text{ m/s}

Step 3: Calculate Additional Distance

Since the time of flight remains unchanged (because vertical motion is unaffected), we use the standard kinematic equation for projectile motion:

t=2viygt = \frac{2 v_{iy}}{g}

t=2(4.1)9.81t = \frac{2(4.1)}{9.81}

t=8.29.810.836 st = \frac{8.2}{9.81} \approx 0.836 \text{ s}

The additional horizontal distance covered due to the velocity increase:

Δx=(vfxvix)t\Delta x = (v_{fx} - v_{ix}) t

Δx=(10.5659.3)(0.836)\Delta x = (10.565 - 9.3)(0.836)

Δx=(1.265)(0.836)\Delta x = (1.265)(0.836)

Δx1.06 m=106 cm\Delta x \approx 1.06 \text{ m} = \mathbf{106} \text{ cm}

Final Answer:

The weights add 106 cm to the jumper’s range.


Would you like a step-by-step breakdown of any part?

Related Questions:

  1. How does the time of flight remain the same in this scenario?
  2. What if the weights were thrown at an angle instead of horizontally?
  3. How would air resistance affect the jumper's range?
  4. What happens if the jumper throws only one weight?
  5. How much additional distance would be gained if the jumper were lighter?

Tip:

Always check whether an external force (like air resistance) might impact conservation of momentum before assuming it's conserved. 🚀

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Math Problem Analysis

Mathematical Concepts

Conservation of Momentum
Projectile Motion
Kinematics

Formulas

Momentum Conservation: m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f
Time of Flight: t = (2 v_{iy}) / g
Horizontal Distance: x = v_x * t

Theorems

Law of Conservation of Momentum
Kinematic Equations of Motion

Suitable Grade Level

Undergraduate Physics / Advanced High School Physics