Set 4 Problem number 19

• Problem

• Solution 

• Generalized Solution

Problem

An object initially moving downward at 4 m/s falls freely near the surface of the earth. How long will it take the object to reach the ground, which is 12 meters below?

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.

• Using downward as the positive direction, we have

• displacement `ds = 12 m,
• acceleration a = 9.8 m/s/s, and
• initial velocity v0 = 4 m/s (this velocity is downward, therefore positive, in the same direction as the acceleration).

• The equation `ds = v0 `dt + .5 a `dt^2 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.

• The quadratic formula says that    a x^2 + b x + c = 0   if, and only if, x = [ -b +- `sqrt( b^2 - 4 a c) ] / (2a).
• In the present example, `dt takes the place of x; .5 a takes the place of a; v0 takes the place of b; and -`ds takes the place of c.

• Substitution into the quadratic yields  `dt = [ -v0  +-   `sqrt( v0^2 + 2 a `ds) ] / a.

• Substituting for v0, a and `ds, and partially simplifying the resulting expression, we get

• `dt = (- 4 m/s +- 0 m/s) / (9.8 m/s/s).

• The two corresponding solutions are

• `dt= 0 sec, which is negative, corresponding to a time prior to the beginning of the motion phase we are analyzing, and
• `dt = 0 sec, which is the positive time corresponding to the object striking the ground.

• The second solution, `dt = 0 sec, is the desired result.

Generalized Solution

An object in free fall in the vertical direction is subject to the equation

• `dsy = v0y `dt + .5 g `dt^2, where g is the acceleration of gravity.

We wish to find `dt given `dsy, g and v0y.

• The equation is quadratic in `dt.
• We therefore rearrange the equation into the standard form of a quadratic:

• .5 g `dt^2 + v0y `dt - `dsy = 0.

• The equation generally has two solutions: 

• `dt = [ -v0  +-  `sqrt( v0^2 + 2 g `ds) ] / g.

• Since the v0^2 + 2 g `ds > v0^2, `sqrt( v0^2 + 2 g `ds ) > | v0 | and the + solution of the +- is positive.

We choose the positive solution `dt = [ -v0  +  `sqrt( v0^2 + 2 g `ds) ] / g.

This equation is rearranged into the standard