Question

A man falls 2.0 m to the floor (assume down to be negative)

how long does the fall take?
what is his velocity when he hits the floor?

Answer 1

Approximately
`0.64\; {\rm s}`.

Velocity: approximately
`6.3\; {\rm m\cdot s^(-1)}`.

(Assuming that air resistance is negligible, and that
`g = 9.81\; {\rm m\cdot s^(-2)}` assuming that the fall started from rest.)

Step-by-step explanation:

If air resistance is negligible, acceleration will be constantly
`a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}` during the fall.

Let
`x` denote the displacement. In this question,
`x = (-2.0)\; {\rm m}`. Note that
`x` is negative since the position after the fall is below the initial position.

The initial velocity at the beginning of the fall is
`u = 0\; {\rm m\cdot s^(-1)}` under the assumptions.

Let
`v` denote the final velocity right before landing. Apply the SUVAT equation
`v^(2) - u^(2) = 2\, a\, x` to find
`v\!`. Rearrange this equation to obtain:

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

Substitute in
`a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}`,
`x = (-2.0)\; {\rm m}`, and
`u = 0\; {\rm m\cdot s^(-1)}` and solve for
`v`

`\begin{aligned}v &= -\sqrt{2\, a\, x + u^(2)} \\ &= -\sqrt{2 \,((-9.81)\; {\rm m\cdot s^(-2)}) \, ((-2.0)\; {\rm m}) + (0\; {\rm m\cdot s^(-1)})^(2)} \\ &= \left(-√(2 \, (-9.81)\, (-2.0))\right)\; {\rm m\cdot s^(-1)} \\ &\approx (-6.26)\; {\rm m\cdot s^(-1)} \\ &\approx (-6.3)\; {\rm m\cdot s^(-1)}\end{aligned}`.

(Negative since the final velocity points downwards to the ground.)

In other words, the velocity right before landing would be approximately
`(-6.3)\; {\rm m\cdot s^(-1)}`.

The change in velocity during the fall would be:

`\begin{aligned}v - u &\approx (-6.26)\; {\rm m\cdot s^(-1)} - (0\; {\rm m\cdot s^(-1)}) = (-6.26)\; {\rm m\cdot s^(-1)}\end{aligned}`.

Divide the change in velocity by acceleration to find the duration
`t` of the fall:

`\begin{aligned}t &= (v - u)/(a) \\ &\approx \frac{(-6.26)\; {\rm m\cdot s^(-1)}}{(-9.81)\; {\rm m\cdot s^(-2)}} \\ &\approx 0.64\; {\rm s}\end{aligned}`.