SOLUTION: A ball is launched into the sky at 54.4 feet per second from a 268.8 meter tall building. The equation for the ball’s height, h, at time t seconds is h = -3.2t^2 + 54.4t + 268.8. W

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: A ball is launched into the sky at 54.4 feet per second from a 268.8 meter tall building. The equation for the ball’s height, h, at time t seconds is h = -3.2t^2 + 54.4t + 268.8. W      Log On


   

Question 1077978: A ball is launched into the sky at 54.4 feet per second from a 268.8 meter tall building. The equation for the ball’s height, h, at time t seconds is h = -3.2t^2 + 54.4t + 268.8. When will the ball strike the ground?

Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A ball is launched into the sky at 54.4 feet per second from a 268.8 meter tall building. The equation for the ball’s height, h, at time t seconds is h = -3.2t^2 + 54.4t + 268.8. When will the ball strike the ground?
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When h = 0.
-3.2t^2 + 54.4t + 268.8 = 0
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Your units don't match.
268.8 meters = 882 feet
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The commonly used function in English units is
h(t) = -16t^2 + 54.4t + 882
Solve for t when h(t) = 0. Ignore the negative solution.
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the -3.2t^2 term might apply to some other planet, but not Earth.
Earth's acceleration due to gravity is 32 ft/sec/sec.

Answer by ikleyn(52779) About Me  (Show Source):
You can put this solution on YOUR website!
.
Your equation is written with a mistake.

The correct writing is
h = -16t^2 + 54.4t + 268.8.

The ball will strike the ground when h = 0, i.e.
-16t^2 + 54.4t + 268.8 = 0.