SOLUTION: A parachutist is 800 feet above the ground when she opens her parachute. She then falls at a constant rate of 5 feet per second. How do I write this out as an equation?

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Question 118250: A parachutist is 800 feet above the ground when she opens her parachute. She then falls at a constant rate of 5 feet per second. How do I write this out as an equation?
Found 2 solutions by stanbon, ankor@dixie-net.com:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
h(t) = -16t^2+800
is the equation of her height at time t.
It is only true for t = 0 as she will
not descend at the rate of 16 ft/sec^2
since she has a parachute.
Unless you know her descent rate with
the parachute I don't know what equation
you will use to indicate here height at
time "t".
================
Cheers,
Stan H.

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A parachutist is 800 feet above the ground when she opens her parachute. She then falls at a constant rate of 5 feet per second. How do I write this out as an equation?
:
You want x = time in seconds and y = feet above the ground
:
Since it falls 5 ft every second, we have -5x feet from 800 ft
y = -5x + 800
:
The graph:
+graph%28+300%2C+200%2C+-20%2C200%2C+-200%2C+900%2C+-5x%2B800%29+
:
You can see that when x = 160 seconds, the height (y) = 0; (160*5 = 800)