"
Set 4 Problem number 14
An object falls freely from rest in the vicinity of the earth's surface.
- How long will it take the object to reach the ground, a distance of .6
meters below its starting point?
In the vicinity of Earth's surface, the object will accelerate at 9.8 m/s/s.
- If we were given the time duration and acceleration we could reason this
problem out in two steps; but with the given information there is no easy way to reason
out the time.
- So we use the formula `ds = v0 * `dt + .5 a `dt ^ 2, with v0=0, a=9.8
m/s/s and `ds= .6 m.
- Since v0 = 0 the formula becomes `ds = .5 a `dt^2.
- Solving for `dt we obtain `dt = `sqrt{2 `ds / a}
- Note that -`sqrt{2 `ds / a} is also a possible solution, but
in this situation we reject the negative `dt as making no sense.
- `dt = `sqrt(2 * .6 m / (9.8 m/s^2) ) = .349 s.
Starting from rest and accelerating at constant rate g we have
- `ds = v0 `dt + .5 a `dt^2 = .5 a `dt^2, since v0 = 0.
We solve `ds = .5 a `dt^2 for `dt and choose the positive value of `dt:
- `dt = `sqrt ( 2 `ds / a ).
"