SOLUTION: An object is propelled vertically upward with an initial velocity of 20 m/s. The distance s (in meters) of the object from the ground after t seconds is given by {{{ s(t) = -4.9t^2

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: An object is propelled vertically upward with an initial velocity of 20 m/s. The distance s (in meters) of the object from the ground after t seconds is given by {{{ s(t) = -4.9t^2      Log On


   



Question 1120524: An object is propelled vertically upward with an initial velocity of 20 m/s. The distance s (in meters) of the object from the ground after t seconds is given by +s%28t%29+=+-4.9t%5E2+%2B+20t+. When will the object be 15 meters above the ground?
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
+s%28t%29+=+-4.9t%5E2+%2B+20t+. When will the object be 15 meters above the ground?
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when s(t) = 20
20+=+-4.9t%5E2+%2B+20t+
4.9t%5E2+-+20t+%2B+20+=+0+
Solve for t.
You get 2 times, ascending & descending.
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Ooooh, 15, not 20.
The method is the same.
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If you solve for s(t) = 100, eg, you don't get a real number solution because the projectile never gets to 100 meters.

Answer by ikleyn(52750) About Me  (Show Source):
You can put this solution on YOUR website!
.
+s%28t%29 = -4.9t%5E2+%2B+20t+. When will the object be 15 meters above the ground?
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I am here to edit the Alan's solution, which is not exactly correct.

when s(t) = 15.


15 = -4.9t%5E2+%2B+20t+

4.9t%5E2+-+20t+%2B+15+=+0+


Solve for t.


You get 2 times, ascending & descending.







Plot y = -4.9t%5E2+%2B+20t (red)  and  y = 15 (green)