To calculate the maximum height the dart would reach when all the elastic potential energy from the slingshot is converted into gravitational potential energy, we can apply the principle of energy conservation. The total mechanical energy remains constant, so the elastic potential energy (EPE) of the slingshot at full stretch will equal the gravitational potential energy (GPE) of the dart at its maximum height.

The relevant energy equations are:

1. Elastic Potential Energy (EPE): $EPE = \frac{1}{2} k x^2$ where:

   • $k$ is the spring constant (elastic constant of the slingshot),
   • $x$ is the stretch or displacement of the slingshot from its equilibrium position.

2. Gravitational Potential Energy (GPE): $GPE = mgh$ where:

   • $m$ is the mass of the dart,
   • $g$ is the acceleration due to gravity (approximately $9.81 \, \text{m/s}^2$),
   • $h$ is the maximum height.

Energy Conservation:

By conservation of mechanical energy, we have: $EPE = GPE$ $\frac{1}{2} k x^2 = mgh$

Solving for height $h$:

$h = \frac{k x^2}{2mg}$

To calculate the height, you would need the values of the spring constant $k$, the displacement $x$, and the mass of the dart $m$.

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Would you like to provide the values for $k$, $x$, and $m$ so we can calculate the height? Alternatively, I can help with any further details or clarifications.

Here are 5 related questions:

1. How does increasing the stretch of the slingshot affect the maximum height?
2. What happens if part of the elastic potential energy is lost due to air resistance?
3. How would the height change if the mass of the dart were doubled?
4. What is the role of the spring constant in determining the height reached?
5. How does gravitational acceleration $g$ vary with altitude, and could this affect the result?

Tip: Always check that energy is fully conserved in the system by accounting for possible losses like air resistance or internal friction.