The linear equation graphed above gives the height in feet above the ground of Shelly t seconds after she opened her parachute when jumping from an airplane. According to the graph, how many seconds after opening her parachute will Shelly be 2,000 feet above the ground?

[tex]\begin{array}{rcl}m & = & \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\ & = & \dfrac{2300 - 2600}{30 - 0}\\\\& = & \dfrac{-300}{30}\\\\& = & \text{-10 ft/s}\\\\\end{array}[/tex]

(b) Locate the y-intercept

The y-intercept is at 2600 ft

(c) Write the equation for the line

h = -10t + 2600

(d) Calculate the time to 2000 ft

[tex]\begin{array}{rcl}h & = & -10t + 2600\\2000 & = & -10t + 2600\\-600 & = & -10t\\t & = & \dfrac{-600}{-10}\\\\& = & \text{60 s}\\\end{array}\\[/tex]

Shelley will be at 2000 ft 60 s after opening the parachute.

codiepienagoya codiepienagoya · Answer 2 · 2021-08-14T17:55:01+02:00

Let the given line pass through the point that is [tex]\bold{(0,2600)\ \ and\ \ (20, 2400)}[/tex]

[tex]\therefore[/tex]

[tex]\to \bold{\frac{H-2400}{t-20}} \bold{= \frac{2600-2400}{0-20}}\\\\[/tex]

[tex]\bold{=\frac{200}{-20} }\\\\ \bold{= -\frac{200}{20}}\\\\ \bold{= - 10}\\\\[/tex]

[tex]\to \bold{H-2400=-10t+200}\\\\\to \bold{H+10t=2400+200}\\\\\to \bold{H+10t=2600}\\\\\to \bold{H=2600-10t}\\\\[/tex]

Let

time (t) in second

Height (h) in feet

for [tex]\bold{\ H=2000\ feet\\}[/tex]

[tex]\to \bold{2000=2600-10t}\\\\\to \bold{10t = 2600- 2000}\\\\\to \bold{10t = 600}\\\\\to \bold{t=\frac{600}{10}}\\\\\to \bold{t= 60\ second}\\\\[/tex]

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