SOLUTION: Skydivers jump out of a plane at an altitude of 3500 meters. Their fall is modeled by the equation h= 3500 - 5t^2where h is the height in meters and t is the number of seconds of

Question 1195705: Skydivers jump out of a plane at an altitude of 3500 meters. Their fall is modeled by the
equation h= 3500 - 5t^2where h is the height in meters and t is the number of seconds of free
fall.
a. How far have the divers fallen after 10 seconds?
b. The divers open their chutes at an altitude of 1000m. How many seconds did they fall?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the equation is h = 3500 - 5t^2
h is the height.
t is the time.

when t = 10, the equation becomes h = 3500 - 5 * 10^2
solve for h to get:
h = 3000.

when the divers open their chutes at a height of 1000 meters, the equation becomes 1000 = 3500 - 5t^2.
add 5t^2 to both sides of the equation and subtract 1000 from both sides of the equation to get:
5t^2 = 3500 - 1000
simplify to get:
5t^2 = 2500
divide both sides of the equation by 5 to get:
t^2 = 500
solve for t to get:
t = sqrt(500) = 22.36067977.

the equation can be graphed.
just replace h with y and t with x to get:
y = 3500 - 5x^2.
y is the height.
x is the time.
the graph is shown below.

let me know if you have any questions.
theo

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