Question

A new ride being built at an amusement park includes a vertical drop of 79.8 meters. Starting from rest, the ride vertically drops that distance before the track curves forward. If friction is neglected, what would be the speed of the roller coaster at the bottom of the drop?

Answer 1

Approximately
`39.6\; {\rm m\cdot s^(-1)}` (assuming that
`g = 9.81\; {\rm m\cdot s^(-2)}`.)

Step-by-step explanation:

Under the assumptions, vertical acceleration of the vehicle during the ride would be equal to the gravitational field strength:
`a = g = 9.81\; {\rm m\cdot s^(-2)}`.

Apply the following SUVAT equation to find the velocity of the vehicle at the bottom of the drop:

`v^(2) - u^(2) = 2\, a\, x`,

Where:

• 
  `v` is the final velocity at the bottom of the drop;
• 
  `u` is the initial velocity at the beginning of the drop;
  `u = 0\; {\rm m\cdot s^(-1)}` since the vehicle started from rest;
• 
  `a = g = 9.81\; {\rm m\cdot s^(-2)}` is the vertical acceleration of the vehicle during the drop;
• 
  `x = 79.8\; {\rm m}` is the vertical displacement of the vehicle during the drop.

Rearrange this equation to find
`v`:

`\begin{aligned}v &= \sqrt{u^(2) + 2\, a\, x} \\ &\approx \sqrt{0^(2) + 2\, (9.81)\, (79.8)} \; {\rm m\cdot s^(-1)} \\ &\approx 39.6\; {\rm m\cdot s^(-1)}\end{aligned}`.

Hence, the speed of this vehicle at the bottom of the drop would be approximately
`39.6\; {\rm m\cdot s^(-1)}`.