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Sagot :
Let's break down the problem step-by-step and solve for the times when the height of the object is 57 feet and when it reaches the ground.
### Given
The height \( h \) of the object after \( t \) seconds is modeled by the equation:
[tex]\[ h = -16t^2 + 62t + 14 \][/tex]
### Part 1: When will the height be 57 feet?
To find the time \( t \) when the height \( h \) is 57 feet, we set up the equation:
[tex]\[ -16t^2 + 62t + 14 = 57 \][/tex]
Next, we rearrange this equation to bring all terms to one side:
[tex]\[ -16t^2 + 62t + 14 - 57 = 0 \][/tex]
[tex]\[ -16t^2 + 62t - 43 = 0 \][/tex]
This is a quadratic equation in the form \( at^2 + bt + c = 0 \) where:
[tex]\[ a = -16, \quad b = 62, \quad c = -43 \][/tex]
Solving this quadratic equation, we get the solutions:
[tex]\[ t = \frac{31}{16} - \frac{\sqrt{273}}{16} \][/tex]
[tex]\[ t = \frac{31}{16} + \frac{\sqrt{273}}{16} \][/tex]
So, the object will be 57 feet high at:
[tex]\[ t = \frac{31 - \sqrt{273}}{16} \quad \text{and} \quad t = \frac{31 + \sqrt{273}}{16} \][/tex]
### Part 2: When will the object reach the ground?
To find the time \( t \) when the object reaches the ground, we set \( h = 0 \) in the height equation:
[tex]\[ -16t^2 + 62t + 14 = 0 \][/tex]
This is already in quadratic form \( at^2 + bt + c = 0 \) where:
[tex]\[ a = -16, \quad b = 62, \quad c = 14 \][/tex]
Solving this quadratic equation, we get the solutions:
[tex]\[ t = \frac{31}{16} - \frac{\sqrt{1185}}{16} \][/tex]
[tex]\[ t = \frac{31}{16} + \frac{\sqrt{1185}}{16} \][/tex]
So, the object will reach the ground at:
[tex]\[ t = \frac{31 - \sqrt{1185}}{16} \quad \text{and} \quad t = \frac{31 + \sqrt{1185}}{16} \][/tex]
### Summary
In conclusion:
- The times when the object will be 57 feet high are:
[tex]\[ t = \frac{31 - \sqrt{273}}{16} \quad \text{seconds} \][/tex]
[tex]\[ t = \frac{31 + \sqrt{273}}{16} \quad \text{seconds} \][/tex]
- The times when the object will reach the ground are:
[tex]\[ t = \frac{31 - \sqrt{1185}}{16} \quad \text{seconds} \][/tex]
[tex]\[ t = \frac{31 + \sqrt{1185}}{16} \quad \text{seconds} \][/tex]
### Given
The height \( h \) of the object after \( t \) seconds is modeled by the equation:
[tex]\[ h = -16t^2 + 62t + 14 \][/tex]
### Part 1: When will the height be 57 feet?
To find the time \( t \) when the height \( h \) is 57 feet, we set up the equation:
[tex]\[ -16t^2 + 62t + 14 = 57 \][/tex]
Next, we rearrange this equation to bring all terms to one side:
[tex]\[ -16t^2 + 62t + 14 - 57 = 0 \][/tex]
[tex]\[ -16t^2 + 62t - 43 = 0 \][/tex]
This is a quadratic equation in the form \( at^2 + bt + c = 0 \) where:
[tex]\[ a = -16, \quad b = 62, \quad c = -43 \][/tex]
Solving this quadratic equation, we get the solutions:
[tex]\[ t = \frac{31}{16} - \frac{\sqrt{273}}{16} \][/tex]
[tex]\[ t = \frac{31}{16} + \frac{\sqrt{273}}{16} \][/tex]
So, the object will be 57 feet high at:
[tex]\[ t = \frac{31 - \sqrt{273}}{16} \quad \text{and} \quad t = \frac{31 + \sqrt{273}}{16} \][/tex]
### Part 2: When will the object reach the ground?
To find the time \( t \) when the object reaches the ground, we set \( h = 0 \) in the height equation:
[tex]\[ -16t^2 + 62t + 14 = 0 \][/tex]
This is already in quadratic form \( at^2 + bt + c = 0 \) where:
[tex]\[ a = -16, \quad b = 62, \quad c = 14 \][/tex]
Solving this quadratic equation, we get the solutions:
[tex]\[ t = \frac{31}{16} - \frac{\sqrt{1185}}{16} \][/tex]
[tex]\[ t = \frac{31}{16} + \frac{\sqrt{1185}}{16} \][/tex]
So, the object will reach the ground at:
[tex]\[ t = \frac{31 - \sqrt{1185}}{16} \quad \text{and} \quad t = \frac{31 + \sqrt{1185}}{16} \][/tex]
### Summary
In conclusion:
- The times when the object will be 57 feet high are:
[tex]\[ t = \frac{31 - \sqrt{273}}{16} \quad \text{seconds} \][/tex]
[tex]\[ t = \frac{31 + \sqrt{273}}{16} \quad \text{seconds} \][/tex]
- The times when the object will reach the ground are:
[tex]\[ t = \frac{31 - \sqrt{1185}}{16} \quad \text{seconds} \][/tex]
[tex]\[ t = \frac{31 + \sqrt{1185}}{16} \quad \text{seconds} \][/tex]
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