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A parachutist jumps from an airplane and immediately opens her parachute. Her altitude in metres, after t seconds, is modelled by the equation y = -11t + 500. A second parachutist jumps 5 secs later and free falls for a few seconds. His altitude during this time is modeled by the equation y = -4.9 (t-4)^2 + 500. When does he catch up to the first parachutist? (HINT: When do they meet in the air?)

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cryssatemp

Answer: They meet in air after 1.6 seconds

Step-by-step explanation:

We have the equations that model the altitude of both parachutist:

[tex]y_{1}=-11t^{2}+500[/tex] (1)

[tex]y_{2}=-4.9(t-4)^{2}+500[/tex] (2)

Both parachutists meet in the air when [tex]y_{1}=y_{2}[/tex]:

[tex]-11t^{2}+500=-4.9(t-4)^{2}+500[/tex] (3)

Rearranging the equation:

[tex]-11t^{2}=-4.9(t^{2}-8t+16)[/tex] (4)

[tex]0=6.1 t^{2}+39.2t-78.4[/tex] (5)

Finding the positive solution for [tex]t[/tex] with the quadratic formula [tex]t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex], where [tex]a=6.1[/tex], [tex]b=39.2[/tex], [tex]c=-78.4[/tex]:

[tex]t=\frac{-39.2\pm\sqrt{39.2^{2}-4(6.1)(-78.4)}}{2(6.1)}[/tex]

[tex]t=1.6 s[/tex]  This is the time when both parachutists meet

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