To determine the speed of the roller coaster at the bottom of the hill, we can use the principle of conservation of mechanical energy. The mechanical energy of a system (ignoring air resistance and friction) is conserved, meaning the total potential energy at the top of the hill will be converted into kinetic energy at the bottom.
First, let’s define the energy forms in this situation:
1. Potential Energy (PE) at the top of the hill:
[tex]\[
PE = m \cdot g \cdot h
\][/tex]
where:
- [tex]\( m \)[/tex] is the mass of the roller coaster (500 kg),
- [tex]\( g \)[/tex] is the acceleration due to gravity (9.8 [tex]\( m/s^2 \)[/tex]),
- [tex]\( h \)[/tex] is the height of the hill (57 m).
2. Kinetic Energy (KE) at the bottom of the hill:
[tex]\[
KE = \frac{1}{2} m v^2
\][/tex]
where:
- [tex]\( m \)[/tex] is the mass of the roller coaster,
- [tex]\( v \)[/tex] is the speed of the roller coaster at the bottom of the hill.
According to the conservation of mechanical energy:
[tex]\[
PE_{top} = KE_{bottom}
\][/tex]
Substitute the formulas for potential energy and kinetic energy:
[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 of the equation:
[tex]\[
g \cdot h = \frac{1}{2} v^2
\][/tex]
Solving for [tex]\( v^2 \)[/tex]:
[tex]\[
2 \cdot g \cdot h = v^2
\][/tex]
Now plug in the values for [tex]\( g \)[/tex] and [tex]\( h \)[/tex]:
[tex]\[
2 \cdot 9.8 \cdot 57 = v^2
\][/tex]
Calculate the right side:
[tex]\[
2 \cdot 9.8 \cdot 57 = 1117.2
\][/tex]
Thus,
[tex]\[
v^2 = 1117.2
\][/tex]
Taking the square root of both sides to solve for [tex]\( v \)[/tex]:
[tex]\[
v = \sqrt{1117.2}
\][/tex]
This yields:
[tex]\[
v \approx 33.4 \, m/s
\][/tex]
Therefore, the speed of the roller coaster at the bottom of the hill is approximately [tex]\( 33.4 \, m/s \)[/tex].
So, the correct answer is:
C. [tex]\( 33.4 \, m/s \)[/tex]
