To find how much higher the starting point of the roller coaster car is compared to the lowest point, we can use the principles of energy conservation. Specifically, we can equate the potential energy at the top of the hill to the kinetic energy at the lowest point.
1. Potential Energy at the Top of the Hill (PE):
The potential energy at the top of the hill is given by:
[tex]\[
PE = m \cdot g \cdot h
\][/tex]
where [tex]\( m \)[/tex] is the mass of the roller coaster car, [tex]\( g \)[/tex] is the acceleration due to gravity, and [tex]\( h \)[/tex] is the height of the hill.
2. Kinetic Energy at the Bottom of the Hill (KE):
The kinetic energy at the bottom of the hill is given by:
[tex]\[
KE = \frac{1}{2} m \cdot v^2
\][/tex]
where [tex]\( m \)[/tex] is the mass of the roller coaster car, and [tex]\( v \)[/tex] is its velocity at the lowest point.
3. Conserving Energy:
According to the conservation of mechanical energy, the potential energy at the top is equal to the kinetic energy at the bottom:
[tex]\[
m \cdot g \cdot h = \frac{1}{2} m \cdot v^2
\][/tex]
4. Solving for Height (h):
By canceling [tex]\( m \)[/tex] from both sides of the equation, we get:
[tex]\[
g \cdot h = \frac{1}{2} v^2
\][/tex]
Rearranging to solve for [tex]\( h \)[/tex]:
[tex]\[
h = \frac{\frac{1}{2} v^2}{g}
\][/tex]
Substituting the given values ([tex]\( v = 26 \, \text{m/s} \)[/tex] and [tex]\( g = 9.80 \, \text{m/s}^2 \)[/tex]):
[tex]\[
h = \frac{\frac{1}{2} \cdot (26)^2}{9.80}
\][/tex]
Simplifying the numerator:
[tex]\[
h = \frac{0.5 \cdot 676}{9.80}
\][/tex]
Continuing the simplification:
[tex]\[
h = \frac{338}{9.80}
\][/tex]
Therefore, the height [tex]\( h \)[/tex] is:
[tex]\[
h \approx 34.49 \, \text{m}
\][/tex]
So, the height difference between the top of the hill and the lowest point of the track is approximately 34 meters. The correct answer is:
34 m
