Question 893279: An object is thrown with a velocity of v, in feet/second, from an initial height of s feet. The object's height at any time,t, in seconds, is given by h= -16t ² + Vt+s.How long would it take an object thrown downward with an initial velocity of 25 ft/s from an initial height of 5280 ft to reach the ground? (Round your answer to the nearest tenth.)
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! An object is thrown with a velocity of v, in feet/second, from an initial height of s feet. The object's height at any time,t, in seconds, is given by h= -16t ² + Vt+s.How long would it take an object thrown downward with an initial velocity of 25 ft/s from an initial height of 5280 ft to reach the ground? (Round your answer to the nearest tenth.)
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h(t) = -16t^2 + Vt + s
h(t) = -16t^2 - 25t + 5280
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It reaches the ground when h(t) = 0
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h(t) = -16t^2 - 25t + 5280 = 0
| Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc) |
Quadratic equation  (in our case  ) has the following solutons:
For these solutions to exist, the discriminant  should not be a negative number.
First, we need to compute the discriminant :  .
Discriminant d=338545 is greater than zero. That means that there are two solutions:  .
Quadratic expression  can be factored:
Again, the answer is: -18.9639437377964, 17.4014437377964.
Here's your graph:
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t = 17.4 seconds
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