To find the maximum height the ball reaches and the time it will take to get there, we need to determine the vertex of the quadratic function \( h(t) = -16t^2 + 48t + 100 \).
The vertex form of a quadratic function \( ax^2 + bx + c \) has its vertex (which gives the maximum height for a downward-opening parabola) at the time \( t \) given by:
\[ t = -\frac{b}{2a} \]
In this case, \( a = -16 \) and \( b = 48 \).
Plugging in these values:
\[ t = -\frac{48}{2 \times -16} = -\frac{48}{-32} = 1.5 \text{ seconds} \]
Now, to find the maximum height, we substitute \( t = 1.5 \) back into the height function \( h(t) \):
\[ h(1.5) = -16(1.5)^2 + 48(1.5) + 100 \]
Calculating \( (1.5)^2 \):
\[ (1.5)^2 = 2.25 \]
Now substituting:
\[ h(1.5) = -16(2.25) + 48(1.5) + 100 \] \[ h(1.5) = -36 + 72 + 100 \] \[ h(1.5) = 136 \text{ feet} \]
Thus, the maximum height the ball reaches is 136 feet and it takes 1.5 seconds to get there.
Summary:
- Maximum Height: 136 feet
- Time to reach maximum height: 1.5 seconds
As for the key feature of the graph that the problem is asking for, the answer is the vertex.
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