Chapter 9.7, Problem 22PPS

a.

To determine

To calculate how long after jumping did parachutist release parachute.

Answer to Problem 22PPS

7 seconds

Explanation of Solution

Given information:

Free fall height equation:

  $y=-4.9x^2+x+360$

Height after releasing parachute equation:

  $y=-4x+142$

Formula used:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

x are roots of quadratic equation

a, b and c are constants

Calculation:

Free fall height equation:

  $y=-4.9x^2+x+360\mathrm{..........}(1)$

Height after releasing parachute equation:

  $y=-4x+142\mathrm{..............}(2)$

Functions are continuous

Then,

Substituting value of $y=-4x+142$in equation (1)

  $-4x+142=-4.9x^2+x+360$

On solving,

  $-4.9x^2+5x+218=0$

  $a=-4.5$

  $b=5$

  $c=218$

Using formula

  $x=\frac{-5\pm \sqrt{5^2-4(-4.9)(218}}{2(-4.9)}$

On solving,

  $x=\frac{5\pm \sqrt{4297.8}}{9.8}$

On solving,

  $x=\frac{5\pm 65.55}{9.8}$

Two values of x

  $x=\frac{5-65.55}{9.8}$ and $x=\frac{5+65.55}{9.8}$

On solving,

  $x=-6\text{ , 7}$

Then, approximate value of x

  $x=7$ seconds

b.

To determine

To calculate height when parashutist release parachute.

Answer to Problem 22PPS

  $y=127$ meter

Explanation of Solution

Given information:

Free fall height equation:

  $y=-4.9x^2+x+360$

Height after releasing parachute equation:

  $y=-4x+142$

Calculation:

Free fall height equation:

  $y=-4.9x^2+x+360\mathrm{......}(1)$

Height after releasing parachute equation:

  $y=-4x+142\mathrm{..............}(2)$

As Substituting $x=7$in equation(1)

Height when parachutist released parachute

  $y=-4.9{(7)}^2+7+360$

On solving,

  $y=127$ meter