Aparachutist 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? )

They meet in air after 1.6 seconds

Step-by-step explanation:

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

(1)

(2)

Both parachutists meet in the air when :

(3)

Rearranging the equation:

(4)

(5)

Finding the positive solution for with the quadratic formula , where , , :

 This is the time when both parachutists meet

hi there!

the answer will be d: 2

to complete this square you have to rotate about the origin 270 degrees clockwise.