SOLUTION: A pair of skydivers jump out of an airplane 5.5 km above the ground. The equation H= 5500-5t2 is an approximate model for the diver's altitude in metres at t seconds after jumping

Question 109097: A pair of skydivers jump out of an airplane 5.5 km above the ground. The equation H= 5500-5t2 is an approximate model for the diver's altitude in metres at t seconds after jumping out of a plane.
a.) After 10 seconds, how far have the divers fallen?
b.) They open their chutes at an altitude of 1000 metres. How long did they free-fall?
c.) If a parachute does not open at 1000 metres, how much time is left to use the emergency chute?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A pair of skydivers jump out of an airplane 5.5 km above the ground. The equation H = 5500-5t^2 is an approximate model for the diver's altitude in metres at t seconds after jumping out of a plane.
:
a.) After 10 seconds, how far have the divers fallen?
:
Substitute 10 for t in 5t^2:
Meters fallen = 5(10^2) = 5(100) = 500 meters after 10 seconds
:
b.) They open their chutes at an altitude of 1000 metres. How long did they free-fall?
5500 - 5t^2 = 1000
-5t^2 = 1000 - 5500
t^2 = ; divided both sides by -5
t^2 = +900
t =
t = 30 seconds
:
c.) If a parachute does not open at 1000 metres, how much time is left to use the emergency chute?
Find how long it would take to hit the ground
5500 - 5t^2 = 0
-5t^2 = -5500
t^2 =
t =
t = 33.2 seconds
:
It looks like he only has about 3 seconds to trip the emergency chute
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