Set 4 Problem number 19

• Problem

• Solution 

• Generalized Solution

Problem

How long will it take an object to reach the ground after it is is given a downward velocity of  9 m/s at an altitude of 10 meters above the ground?

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 = 10 m,
• acceleration a = 9.8 m/s/s, and
• initial velocity v0 = 9 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 = (- 9 m/s +- 16.64 m/s) / (9.8 m/s/s).

• The two corresponding solutions are

• `dt=-2.617 sec, which is negative, corresponding to a time prior to the beginning of the motion phase we are analyzing, and
• `dt = .7795918 sec, which is the positive time corresponding to the object striking the ground.

• The second solution, `dt = .7795918 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