Problem: 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 71 meters below its starting point?
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Solution: In the vicinity of Earth's surface, the object will accelerate at 9.8 m/s/s. If we were given time 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 s= 71 m. We obtain the equation
71 m = (0 m/s) *`dt + .5 (9.8 m/s/s) * `dt ^ 2, or 71 m / s = (4.9 m/s/s)'dt ^ 2.
Solving for `dt we obtain `dt = `sqrt{( 71 m/s)/4.9 m/s/s)} = 3.806 s.
We put the equation into standard quadratic form to obtain
.5 a `dt^2 + v0 `dt - `ds = 0, a quadratic in `dt.
Solving for `dt we get
`dt = [ -v0 +- `sqrt( v0^2 - 4 (.5 a) (-`ds) ) ] / ( 2 * .5 a)
= -v0 / a +- `sqrt ( v0^2 + 2 a `ds) / a.
The - of the +- clearly gives a negative `dt for any positive v0, and is not relevant here. The + gives a positive `dt, since `sqrt ( v0^2 + 2 a `ds) / a > v0 / a, and gives us the time required to reach the ground.
Since in the present example v0 = 0, the expression for `dt simplifies to `sqrt( 2 `ds / a ).
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