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?

   Hint (a):

   Substitute $0$ in for $x$.

   Answer (a):

   $y = − 16(0^2) + 64(0) + 80 = 80$

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.

   Hint (b):

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

   Answer (b):

   $y = − 16(3^2) + 64(3) + 80$
   $y = − 144 + 192 + 80$
   $y = − 144 + 272$
   $y = 128$ ft

   $y = − 16(0.5^2) + 64(0.5) + 80$
   $y = − 4 + 32 + 80$
   $y = − 4 + 112$
   $y = 10$8 ft

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

   Hint (c):

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

   Answer (c):

   $−16x^2 + 64x + 80 = 0$
   $−16(x^2 − 4x − 5)$
   $−16(x − 5)(x + 1)$
   $x = 5$ seconds