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12-87.

Perry threw a tennis ball up into the air from the edge of a cliff. The height of the ball was $y=-16x^2+64x+80$, where $y$ represents the height in feet of the ball above ground at the bottom of the cliff, and $x$ represents the time in seconds after the ball is thrown.

1. How high was the ball when it was thrown? How do you know?

Substitute $0$ in for $x$.

$y=-16\left(0^2\right)+64\left(0\right)+80=80$

2. What was the height of the ball $3$ seconds after it was thrown? What was its height $\frac { 1 } { 2 }$ a second after it was thrown? Show all work.

Use the same method you used to solve part (a).

$\begin{array}{l} y = - 16(3^2) + 64(3) + 80 \\ y = - 144 + 192 + 80 \\ y = - 144 + 272 \\ y = 128 \; \text{ft} \end{array}$

$\begin{array}{l} y = - 16(0.5^2) + 64(0.5) + 80 \\ y = - 4 + 32 + 80 \\ y = - 4 + 112 \\ y = 108 \; \text{ft} \end{array}$

3. When did the ball hit the ground? Write and solve an equation that represents this situation.

If the ball hits the ground, what is its height above the ground?

$\begin{array}{l} -16x^2 + 64x + 80 = 0 \\ -16 (x^2 - 4x - 5) \\ -16 (x - 5)(x+1) \\ x = 5 \; \text{seconds} \end{array}$