Math Problem Statement

A javelin thrower standing at rest holds the center of the javelin behind her head, then accelerates it through a distance of 70 cm as she throws. She releases the 600 g javelin 2.0 m above the ground traveling at an angle of 30∘ above the horizontal. Top-rated javelin throwers do throw at about a 30∘ angle, not the 45∘ you might have expected, because the biomechanics of the arm allow them to throw the javelin much faster at 30∘ than they would be able to at 45∘. In this throw, the javelin hits the ground 62 m away. What was the acceleration of the javelin during the throw? Assume that it has a constant acceleration.

Solution

To find the acceleration of the javelin during the throw, we will use kinematic equations and some physics principles. Here’s the step-by-step approach:

Step 1: Key Information and Assumptions

  • Distance of throw: 62m62 \, \text{m}
  • Launch angle: 3030^\circ
  • Release height: 2.0m2.0 \, \text{m}
  • Mass of javelin: 600g=0.6kg600 \, \text{g} = 0.6 \, \text{kg}
  • Acceleration is constant while the javelin is in the thrower’s hand.
  • Distance over which acceleration happens: 70cm=0.70m70 \, \text{cm} = 0.70 \, \text{m}
  • We need to find the acceleration during the throw.

Step 2: Decompose the Motion

When the javelin is thrown, its motion can be decomposed into two components:

  1. Horizontal component of velocity: v0x=v0cos30v_{0x} = v_0 \cos 30^\circ
  2. Vertical component of velocity: v0y=v0sin30v_{0y} = v_0 \sin 30^\circ

Here, v0v_0 is the initial speed of the javelin when it is released. To find v0v_0, we’ll use projectile motion principles.

Step 3: Time of Flight (using vertical motion)

Since the javelin is launched 2.0 meters above the ground and its vertical motion is subject to gravity, we can calculate the time of flight using the following equation of motion for vertical displacement:

y=v0yt12gt2y = v_{0y} t - \frac{1}{2} g t^2 where:

  • y=2.0my = -2.0 \, \text{m} (since the javelin ends 2.0 m lower than it starts),
  • v0y=v0sin30v_{0y} = v_0 \sin 30^\circ,
  • g=9.8m/s2g = 9.8 \, \text{m/s}^2,
  • tt is the total time of flight.

We also know that the horizontal range of the projectile is given by:

x=v0xtx = v_{0x} t where:

  • x=62mx = 62 \, \text{m},
  • v0x=v0cos30v_{0x} = v_0 \cos 30^\circ.

Step 4: Solve for the Initial Velocity v0v_0

Let me calculate the exact value of v0v_0.The positive solution for the initial velocity v0v_0 is approximately 25.78m/s25.78 \, \text{m/s}, and the time of flight tt is about 2.78seconds2.78 \, \text{seconds}.

Step 5: Calculate the Acceleration During the Throw

Now that we know the initial velocity v0=25.78m/sv_0 = 25.78 \, \text{m/s}, we can use the following kinematic equation to find the constant acceleration aa during the throw:

v02=2adv_0^2 = 2 a d where:

  • v0=25.78m/sv_0 = 25.78 \, \text{m/s},
  • d=0.70md = 0.70 \, \text{m} (the distance over which the javelin was accelerated),
  • aa is the acceleration.

Let me calculate the acceleration.The acceleration of the javelin during the throw is approximately 474.72m/s2474.72 \, \text{m/s}^2.

Would you like further details on this, or do you have any other questions?

Here are 5 related questions to expand your understanding:

  1. How does the angle of release affect the range of a projectile?
  2. What is the significance of constant acceleration in projectile motion?
  3. How would air resistance affect the javelin's trajectory and acceleration?
  4. What is the relationship between force and acceleration in this scenario?
  5. How could a thrower increase the range of a javelin?

Tip: The optimal angle for maximum range isn't always 45° when height differences are involved in projectile motion.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Projectile Motion
Trigonometry

Formulas

y = v_{0y} t - (1/2) g t^2
x = v_{0x} t
v_0^2 = 2 a d

Theorems

Equations of motion for uniformly accelerated motion
Decomposition of projectile motion into horizontal and vertical components

Suitable Grade Level

Grade 10-12 (High School Physics)