Let's solve this problem step-by-step.
### Step 1: Calculate the Potential Energy at the Top of the Hill
The potential energy (PE) at the top of the hill can be calculated using the formula:
[tex]\[ \text{PE} = m \cdot g \cdot h \][/tex]
where:
- [tex]\( m = 250 \, \text{kg} \)[/tex] is the mass of the roller coaster,
- [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex] is the acceleration due to gravity,
- [tex]\( h = 88 \, \text{m} \)[/tex] is the height of the hill.
Plugging in the values, we get:
[tex]\[ \text{PE} = 250 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 88 \, \text{m} \][/tex]
[tex]\[ \text{PE} = 215600 \, \text{J} \][/tex]
So, the potential energy at the top of the hill is [tex]\( 215600 \, \text{J} \)[/tex].
### Step 2: Understand Energy Conversion
At the bottom of the hill, all of the potential energy will have been converted into kinetic energy (assuming no energy is lost due to friction or air resistance). Therefore, at the bottom, the potential energy equals the kinetic energy (KE):
[tex]\[ \text{KE} = \text{PE} \][/tex]
### Step 3: Relate Kinetic Energy to Velocity
The kinetic energy of a mass [tex]\( m \)[/tex] moving at a velocity [tex]\( v \)[/tex] is given by:
[tex]\[ \text{KE} = \frac{1}{2} m v^2 \][/tex]
### Step 4: Set Up the Equation
We know the kinetic energy:
[tex]\[ 215600 \, \text{J} = \frac{1}{2} \times 250 \, \text{kg} \times v^2 \][/tex]
### Step 5: Solve for [tex]\( v \)[/tex]
First, multiply both sides by 2 to isolate [tex]\( v^2 \)[/tex]:
[tex]\[ 2 \times 215600 \, \text{J} = 250 \, \text{kg} \times v^2 \][/tex]
[tex]\[ 431200 = 250 v^2 \][/tex]
Next, divide both sides by the mass:
[tex]\[ v^2 = \frac{431200}{250} \][/tex]
[tex]\[ v^2 = 1724.8 \][/tex]
Finally, take the square root of both sides to find [tex]\( v \)[/tex]:
[tex]\[ v = \sqrt{1724.8} \][/tex]
[tex]\[ v = 41.53 \, \text{m/s} \][/tex]
Thus, the roller coaster is traveling at approximately [tex]\( 41.53 \, \text{m/s} \)[/tex] when it reaches the bottom of the hill.
