Question

A roller coaster has a mass of 450 kg. It sits at the top of a hill with height 49

m. If it drops from this hill, how fast is it going when it reaches the bottom?

Answer 1

Approximately
`31\; {\rm m\cdot s^(-1)}` assuming that there is no friction on the rollercoaster, and that
`g = 9.81\; {\rm m\cdot s^(-2)}`.

Step-by-step explanation:

Make use of the SUVAT equation
`v^(2) - u^(2) = 2\, a\, x`, where:

• 
  `v` is the final velocity of the moving object,
• 
  `u` is the initial velocity of the object,
• 
  `a` is the acceleration of the object, and
• 
  `x` is the displacement of the object.

If there is no friction on the rollercoaster, the acceleration of the rollercoaster would be equal in magnitude to the gravitational field strength,
`g`:

`a = g = 9.81\; {\rm m \cdot s^(-2)}`.

The initial velocity of this rollercoaster is
`0\; {\rm m\cdot s^(-1)}` (that is,
`u = 0\; {\rm m\cdot s^(-1)}`) since the rollercoaster was initially stationary. The displacement of this rollercoaster would be
`x = 49\; {\rm m}` (same as the height of the hill.)

Rearrange the equation
`v^(2) - u^(2) = 2\, a\, x` to find an expression for the final velocity
`v` of this rollercoaster:

`\begin{aligned}v &= \sqrt{2\, a\, x + u^(2)} \\ &= \sqrt{2 * 9.81\; {\rm m\cdot s^(-2) * 49\; {\rm m} + (0\; \rm m\cdot s^(-1)})^(2)}\\ &\approx 31\; {\rm m\cdot s^(-1)}\end{aligned}`.