What is the average speed of a projectile when thrown from sling in an overhead manner Assumptions weighs 100 grams length arm 29 inches tip fingers to pit 20 in?

Given:

- Mass of the projectile, $m = 100\ \text{g} = 0.1 \ \text{kg}$

- Length of the arm, $L = 29 \ \text{in} = 0.7366 \ \text{m}$

- Distance from the tip of the fingers to the pit, $r = 20 \ \text{in} = 0.508 \ \text{m}$

To find:

- Average speed of the projectile, $v_{avg}$

Solution:

The average speed of the projectile can be found using the formula:

$$v_{avg} = \frac{\Delta x}{\Delta t}$$

Where,

- $\Delta x$ is the displacement of the projectile, and

- $\Delta t$ is the time taken by the projectile to cover this displacement.

First, we need to find the displacement of the projectile. The displacement is the distance between the initial and final positions of the projectile. In this case, the initial position of the projectile is at the tip of the fingers, and the final position is at the pit. Therefore, the displacement is:

$$\Delta x = r = 0.508 \ \text{m}$$

Next, we need to find the time taken by the projectile to cover this displacement. The time taken can be found using the formula:

$$\Delta t = \frac{2L}{v}$$

Where,

- $v$ is the velocity of the projectile.

The velocity of the projectile can be found using the formula:

$$v = \sqrt{2gL}$$

Where,

- $g$ is the acceleration due to gravity ($g = 9.8 \ \text{m/s}^2$).

Substituting the values of $L$ and $g$ into the formula, we get:

$$v = \sqrt{2(9.8 \ \text{m/s}^2)(0.7366 \ \text{m})} = 4.13 \ \text{m/s}$$

Now, we can substitute the values of $\Delta x$ and $\Delta t$ into the formula for average speed:

$$v_{avg} = \frac{0.508 \ \text{m}}{\frac{2(0.7366 \ \text{m})}{4.13 \ \text{m/s}}} = 2.81 \ \text{m/s}$$

Therefore, the average speed of the projectile is $2.81 \ \text{m/s}$.