To determine the velocity of the roller coaster when it reaches the bottom of the drop, we can use the principles of energy conservation, specifically the conservation of gravitational potential energy and kinetic energy.
1. Understanding the Problem:
- The roller coaster has a mass of 500 kg.
- It drops from rest, so its initial velocity ([tex]\(v_0\)[/tex]) is 0 m/s.
- The height of the drop is 57 meters.
- The acceleration due to gravity ([tex]\(g\)[/tex]) is 9.8 m/s[tex]\(^2\)[/tex].
2. Calculating Initial Gravitational Potential Energy (GPE):
- GPE at the top is given by the formula [tex]\( \text{GPE} = m \cdot g \cdot h \)[/tex], where [tex]\(m\)[/tex] is mass, [tex]\(g\)[/tex] is acceleration due to gravity, and [tex]\(h\)[/tex] is height.
- [tex]\[ \text{GPE}_{\text{top}} = 500 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \cdot 57 \, \text{m} \][/tex]
3. Transforming Potential Energy to Kinetic Energy (KE):
- At the bottom, as the roller coaster has come down, all the gravitational potential energy is converted into kinetic energy.
- KE at the bottom is given by the formula [tex]\( \text{KE} = \frac{1}{2} m v^2 \)[/tex].
4. Set GPE Equal to KE to Solve for Velocity (v):
- Since [tex]\( \text{GPE}_{\text{top}} = \text{KE}_{\text{bottom}} \)[/tex]:
- [tex]\[ m \cdot g \cdot h = \frac{1}{2} m v^2 \][/tex]
- We can cancel out the mass ([tex]\(m\)[/tex]) from both sides:
- [tex]\[ g \cdot h = \frac{1}{2} v^2 \][/tex]
- Solve for [tex]\(v\)[/tex]:
- [tex]\[ v^2 = 2 \cdot g \cdot h \][/tex]
- [tex]\[ v = \sqrt{2 \cdot g \cdot h} \][/tex]
5. Substitute the Known Values:
- [tex]\[ v = \sqrt{2 \cdot 9.8 \, \text{m/s}^2 \cdot 57 \, \text{m}} \][/tex]
6. Calculate the Value:
- [tex]\[ v = \sqrt{2 \cdot 9.8 \cdot 57} \][/tex]
- [tex]\[ v = \sqrt{1117.2} \][/tex]
- [tex]\[ v \approx 33.42 \, \text{m/s} \][/tex]
So, the velocity at the bottom of the drop is approximately [tex]\(33.4 \, \text{m/s}\)[/tex], which corresponds to option A.
Therefore, the correct answer is:
A. [tex]\(33.4 \, \text{m/s}\)[/tex].
