Problem: An object initially moving downward at 5.2 m/s falls freely near the surface of the earth. How long will it take the object to reach the ground, which is 13 meters below?
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Solution: The initial velocity, acceleration and downward displacement are given; the most appropriate solution strategy is to use the equation `ds = v0 * t + .5 a `dt ^ 2. Here, using downward as the positive direction, we have displacement s = 13 m, acceleration a = 9.8 m/s/s, and initial velocity v0 = 5.2m/s (downward, therefore positive, in the same direction as a). The equation relating `ds, a and `dt can be written in the standard form of a quadratic equation as .5 a `dt ^ 2 + v0 * t - `ds =0. Substitution and partial simplification yield `dt = (- 5.2 m/s +- 281.8 m/s) / (9.8 m/s/s). The two corresponding solutions are t= 2.243 sec, which is negative, corresponding to a time prior to the beginning of the motion phase we are analyzing, and `dt = 1.182 sec, which is the positive time corresponding to the object striking the ground.
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.
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