SOLUTION: A skydiver jumps from an altitude of 4000 m. Her height above the ground in metres after t seconds is given by the function h = -5t + 4000. She free falls until she is 875 m above

Question 1106910: A skydiver jumps from an altitude of 4000 m. Her height above the ground in metres after t seconds is given by the function h = -5t + 4000. She free falls until she is 875 m above the ground, then pulls her chute. How many seconds is her free fall? (Solve algebraically, not by graphing please)
Found 2 solutions by addingup, ikleyn:
Answer by addingup(3677)   (Show Source): You can put this solution on YOUR website!
-5t + 4000 = h
-5t + 4000 = 4000-875 = 3125
-5t = -875
t = -875/-5 = 175 seconds
Note: this is a heck of a free fall. In real life the free fall lasts less than 60 seconds.

Answer by ikleyn(52887)   (Show Source): You can put this solution on YOUR website!
.
A skydiver jumps from an altitude of 4000 m. Her height above the ground in metres after t seconds is given by the function h = -5t + 4000.
She free falls until she is 875 m above the ground, then pulls her chute. How many seconds is her free fall?
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You INCORRECTLY presented your function.

The correct form is h = -5t^2 + 4000,     when the unit of length/distance the"meter" is used.

It is the   STANDARD FORM   for the height   at FREE FALL.

To answer the question, you need to solve the quadratic equation -5*t^2 + 4000 = 875. ====> 5t^2 = 4000-875 = 3125 ====> t^2 = = 625 ====> t = = 25. Answer. 25 seconds.

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Be aware ! The solution by @addingup is   I N C O R R E C T !.

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