To write the equation of the height of the ball above ground level as a function of time \( t \), we can use the standard form of a quadratic equation:
\[ h(t) = at^2 + bt + c \]
where:
- \( h(t) \) is the height above ground,
- \( a \), \( b \), and \( c \) are constants,
- \( t \) is the time in seconds.
Given that the height of the ball is 48 feet when it is thrown, we have \( c = 48 \) (the height of the building).
We are also given two additional pieces of information:
- At \( t = 1 \) second, the ball is 64 feet above ground.
- At \( t = 3 \) seconds, the ball reaches ground level (0 feet).
Using the height at \( t = 1 \): \[ h(1) = a(1)^2 + b(1) + 48 = 64 \] This simplifies to: \[ a + b + 48 = 64 \implies a + b = 16 \quad \text{(Equation 1)} \]
Using the height at \( t = 3 \): \[ h(3) = a(3)^2 + b(3) + 48 = 0 \] This simplifies to: \[ 9a + 3b + 48 = 0 \implies 9a + 3b = -48 \implies 3a + b = -16 \quad \text{(Equation 2)} \]
Now we have a system of equations:
- \( a + b = 16 \)
- \( 3a + b = -16 \)
We can solve this system by substituting \( b \) from Equation 1 into Equation 2.
From Equation 1: \[ b = 16 - a \]
Substituting into Equation 2: \[ 3a + (16 - a) = -16 \] This simplifies to: \[ 3a + 16 - a = -16 \] \[ 2a + 16 = -16 \] \[ 2a = -32 \implies a = -16 \]
Now substitute \( a \) back into Equation 1 to find \( b \): \[ -16 + b = 16 \implies b = 32 \]
Having found \( a \) and \( b \), we plug these values back into the general equation: \[ h(t) = -16t^2 + 32t + 48 \]
Thus, the equation of the height above ground level as a function of time \( t \) is:
\[ \boxed{-16t^2 + 32t + 48} \]
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