Answer :
Answer:
The maximum speed of the car at the bottom of that drop is 26.34 m/s.
Explanation:
Given that,
The maximum vertical distance covered by the roller coaster, h = 35.4 m
We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :
[tex]mgh=\dfrac{1}{2}mv^2[/tex]
[tex]v=\sqrt{2gh}[/tex]
[tex]v=\sqrt{2\times 9.8\times 35.4}[/tex]
v = 26.34 m/s
So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.
The maximum speed of the car at the bottom of the drop is 26.34 m/s.
The maximum speed of the car can be determined using the equation below.
Formula:
- v² = u²+2gs..................... Equation 1
Where:
- v = maximum speed of the car at the bottom of the drop.
- u = Initial speed of the car
- s = maximum vertical distance
- g = acceleration due to gravity.
From the question,
Given:
- s = 35.4 m
- g = 9.8 m/s²
- u = 0 m/s (drop from a height)
Substitute these values into the equation
- v² = 0²+(2×35.4×9.8)
- v² = 693.84
- v = √(693.84)
- v = 26.34 m/s
Hence, the maximum speed of the car at the bottom of the drop is 26.34 m/s
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