A roller coaster starts at rest from a height of 110m, and accelerates down the track to a height of 10m. Find the speed it can reach, assuming no friction.
\(v^{2}\) = \(u^{2}\) + 2ar
\(v^{2}\) = 0 + 2 9.8 m/s (110 m - 10m)
v = \(\sqrt{1960}\) = 44 m/s
if it is going down vertically. if there is some angle - it would be another story
Answer: The final velocity of the roller coaster is 44.3 m/s
Explanation:
To calculate the final velocity of the roller coaster, we use third equation of motion:
\(v^2-u^2=2as\)
where,
s = distance traveled = 110 - 10 = 100 m
u = initial velocity of the roller coaster = 0 m/s
v = final velocity of the roller coaster = ? m/s
a = acceleration due to gravity = \(9.8m/s^2\)
Putting values in above equation, we get:
\(v^2-(0)^2=2\times 9.8\times 100\\\\v=\sqrt{1960}=44.3m/s\)
Hence, the final velocity of the roller coaster is 44.3 m/s