To solve the problem, we need to analyze the function \( h(t) = -16t^2 + 56t \) and find the vertex, as well as understand the domain in the context of the problem.
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Finding the time to reach maximum height (Vertex): The vertex \( t \) of a parabolic function in the form \( h(t) = at^2 + bt + c \) can be found using the formula \[ t = -\frac{b}{2a}. \] Here, \( a = -16 \) and \( b = 56 \).
Plugging in the values: \[ t = -\frac{56}{2 \times -16} = -\frac{56}{-32} = \frac{56}{32} = 1.75 \text{ seconds}. \]
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Finding the maximum height: To find the maximum height, plug \( t = 1.75 \) back into the height function \( h(t) \): \[ h(1.75) = -16(1.75)^2 + 56(1.75). \] Calculating \( (1.75)^2 = 3.0625 \): \[ h(1.75) = -16(3.0625) + 56(1.75) = -49 + 98 = 49 \text{ feet}. \]
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Domain considerations: The ball is kicked from the ground, so \( h(t) \) must be greater than or equal to 0. Setting \( h(t) \) to 0 gives: \[ -16t^2 + 56t = 0. \] Factoring out \( t \): \[ t(-16t + 56) = 0 \implies t = 0 \text{ or } t = \frac{56}{16} = 3.5. \] This means the ball will hit the ground again at \( t = 3.5 \) seconds.
Putting it all together, we have:
- The ball took 1.75 seconds to reach its maximum height of 49 feet.
- The domain in context to this situation is 0 to 3.5 because the ball hits the ground at 3.5 seconds.
So, the compiled responses would be:
- Response area seconds to reach its maximum height: 1.75
- Response area feet: 49
- Response area: hits the ground at 3.5 seconds.
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