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: $62 \, \text{m}$
• Launch angle: $30^\circ$
• Release height: $2.0 \, \text{m}$
• Mass of javelin: $600 \, \text{g} = 0.6 \, \text{kg}$
• Acceleration is constant while the javelin is in the thrower’s hand.
• Distance over which acceleration happens: $70 \, \text{cm} = 0.70 \, \text{m}$
• 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: $v_{0x} = v_0 \cos 30^\circ$
2. Vertical component of velocity: $v_{0y} = v_0 \sin 30^\circ$

Here, $v_0$ is the initial speed of the javelin when it is released. To find $v_0$, 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:

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

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

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

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

• $x = 62 \, \text{m}$,
• $v_{0x} = v_0 \cos 30^\circ$.

Step 4: Solve for the Initial Velocity $v_0$

Let me calculate the exact value of $v_0$.The positive solution for the initial velocity $v_0$ is approximately $25.78 \, \text{m/s}$, and the time of flight $t$ is about $2.78 \, \text{seconds}$.

Step 5: Calculate the Acceleration During the Throw

Now that we know the initial velocity $v_0 = 25.78 \, \text{m/s}$, we can use the following kinematic equation to find the constant acceleration $a$ during the throw:

$v_0^2 = 2 a d$ where:

• $v_0 = 25.78 \, \text{m/s}$,
• $d = 0.70 \, \text{m}$ (the distance over which the javelin was accelerated),
• $a$ is the acceleration.

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

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.