The speed at which the car leaves the track is approximately \(7.14 \, \text{m/s}\).
To find the speed at which the car leaves the track, we can use the principle of conservation of energy. The initial potential energy at the top of the hill is converted into kinetic energy at the bottom, and some of that kinetic energy is then converted back into potential energy at the end of the ramp.
The potential energy at the top of the hill is given by:
\[ PE_{\text{initial}} = m \cdot g \cdot h_{\text{initial}} \]
where:
- \( m \) is the mass of the car,
- \( g \) is the acceleration due to gravity,
- \( h_{\text{initial}} \) is the initial height of the hill.
The kinetic energy at the bottom of the hill is given by:
\[ KE_{\text{bottom}} = \frac{1}{2} m \cdot v_{\text{bottom}}^2 \]
where:
- \( v_{\text{bottom}} \) is the speed of the car at the bottom of the hill.
At the end of the ramp, the height is converted into potential energy again:
\[ PE_{\text{final}} = m \cdot g \cdot h_{\text{final}} \]
where:
- \( h_{\text{final}} \) is the height at the end of the ramp.
Setting the initial potential energy equal to the sum of the kinetic energy and the final potential energy:
\[ PE_{\text{initial}} = KE_{\text{bottom}} + PE_{\text{final}} \]
\[ m \cdot g \cdot h_{\text{initial}} = \frac{1}{2} m \cdot v_{\text{bottom}}^2 + m \cdot g \cdot h_{\text{final}} \]
Now, solve for \(v_{\text{bottom}}\):
\[ v_{\text{bottom}} = \sqrt{2 \cdot g \cdot (h_{\text{initial}} - h_{\text{final}})} \]
Given that \(h_{\text{initial}} = 2.3 \, \text{m}\), \(h_{\text{final}} = 0.91 \, \text{m}\), and \(g = 9.81 \, \text{m/s}^2\), substitute these values into the formula:
\[ v_{\text{bottom}} = \sqrt{2 \cdot 9.81 \cdot (2.3 - 0.91)} \]
\[ v_{\text{bottom}} \approx 7.14 \, \text{m/s} \]
Therefore, the speed at which the car leaves the track is approximately \(7.14 \, \text{m/s}\).