To find the acceleration of the javelin during the throw, we will use kinematic equations. We know that the thrower accelerated the javelin through a distance of $70 \, \text{cm} = 0.70 \, \text{m}$ before release. Given the final speed ($v$) at release, we will first find that speed using projectile motion equations, and then use it to determine acceleration.

Step 1: Determine the initial velocity of the javelin at release.

1. Horizontal motion:

   The horizontal distance covered is $50 \, \text{m}$.

   The horizontal component of the velocity at release is: $v_x = v \cos \theta$

   The time of flight ($t$) is given by: $t = \frac{\text{horizontal distance}}{v_x} = \frac{50}{v \cos 30^\circ}$

2. Vertical motion:

   Using the vertical displacement equation (upward is positive): $y = v_y t - \frac{1}{2} g t^2$ Here, $y = -2 \, \text{m}$ (since the javelin falls 2 m from the release point), $v_y = v \sin 30^\circ$, and $g = 9.8 \, \text{m/s}^2$.

   Substituting $t = \frac{50}{v \cos 30^\circ}$: $-2 = v \sin 30^\circ \left(\frac{50}{v \cos 30^\circ}\right) - \frac{1}{2} \cdot 9.8 \left(\frac{50}{v \cos 30^\circ}\right)^2$

   Solving this equation will give us the value of $v$.

Step 2: Find the acceleration using the kinematic equation.

After finding $v$, we use the kinematic equation: $v^2 = u^2 + 2a s$ where:

• $u = 0$ (initial velocity, as the javelin starts from rest),
• $v$ = final velocity at release,
• $s = 0.70 \, \text{m}$ (distance over which acceleration occurs),
• $a$ = acceleration (unknown).

Rearranging for acceleration: $a = \frac{v^2}{2s}$

Let's perform the calculations to find $v$ and then $a$.The negative value for the release velocity indicates a direction issue in our initial assumption, but we'll focus on the magnitude:

1. Release Velocity ($v$): The magnitude is approximately $23.0 \, \text{m/s}$.

2. Acceleration ($a$): The calculated acceleration is approximately $378 \, \text{m/s}^2$.

This large value of acceleration is due to the javelin being accelerated over a short distance of 0.70 meters.

Would you like more details or have any questions about this? Here are some related questions:

1. How does changing the release angle affect the distance the javelin travels?
2. What is the maximum height reached by the javelin during its flight?
3. How much time does the javelin spend ascending and descending?
4. How would a different acceleration distance affect the release speed?
5. What is the role of initial speed in determining the range of projectile motion?

Tip: Always check the sign convention in projectile motion problems to avoid confusion with velocity directions.