Question

A theme park is planning out a new free-fall ride. The drop is almost perfectly frictionless, with a distance of 190 meters. Assuming that the initial velocity was zero, what would be the speed at the bottom of the drop?

a-53 m/s

b-61 m/s

c-67 m/s

d-72 m/s

Answer 1

Approximately
`(-61)\; {\rm m\cdot s^(-2)}`, assuming that
`g = 9.81\; {\rm m\cdot s^(-2)}`.

Step-by-step explanation:

Under the assumptions, the vehicle would be in a free fall. Acceleration would be constant:
`a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}`.

Let
`u` denote the initial velocity of the vehicle. Let
`v` denote the velocity of the vehicle at the bottom of the drop. Let
`x` denote the displacement of the vehicle during the drop.

It is given that the initial velocity is
`u = 0\; {\rm m\cdot s^(-1)}`. During the drop, displacement would be
`x = (-190)\; {\rm m}` (negative since the vehicle is below where it started.) The value of final velocity
`v` needs to be found.

It is known that the vehicle is moving downwards at the end of the fall. Therefore, the value of
`v\!` would be negative. Apply the SUVAT equation
`v^(2) - u^(2) = 2\, a\, x` to find
`v` from
`u`,
`a`, and
`x`.

`\begin{aligned}v^(2) &= u^(2) + 2\, a\, x\end{aligned}`.

`\begin{aligned}v &= -\sqrt{u^(2) + 2\, a\, x} \\ &= -\sqrt{(0\; {\rm m\cdot s^(-1)})^(2) + 2\, (-9.81\; {\rm m\cdot s^(-2)})\, (-190\; {\rm m})} \\ &\approx (-61)\; {\rm m\cdot s^(-1)}\end{aligned}`.

(Note that
`v` is negative.)

In other words, the velocity of the vehicle would be approximately
`(-61)\; {\rm m\cdot s^(-1)}` at the end of the drop.