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The problem states that your cousin hit a golf ball with an initial velocity of \( 27 \) feet per second, and the height \( h \) of the ball after \( t \) seconds is modeled by the equation:

\[
h = -6t^2 + 27t
\]

### Explanation of Terms
- **Initial velocity (27 feet per second)**: This is the speed at which the ball was initially hit upwards.
- **Equation**: The formula \( h = -6t^2 + 27t \) is a quadratic equation, where:
  - \( -6t^2 \) represents the downward pull of gravity (the negative coefficient shows that gravity reduces the height over time).
  - \( 27t \) represents the initial upward velocity’s effect on the height.

### Potential Questions You Might Be Asked
1. **Maximum Height**: When will the ball reach its maximum height, and what is that height?
2. **Time of Flight**: How long will the ball remain in the air before hitting the ground?
3. **Height at a Specific Time**: What is the height of the ball at a specific time, like \( t = 2 \) seconds?
4. **Vertex of the Parabola**: Where is the vertex of this parabola, and why does it represent the maximum height?
5. **Effect of Changing Initial Velocity**: How would a different initial velocity (like 30 feet per second) affect the height equation?

### Tip
For projectile motion problems like this one, the time to reach maximum height can be found using the formula \( t = \frac{-b}{2a} \), where \( a \) and \( b \) are coefficients from the quadratic equation \( at^2 + bt + c \).

Your cousin hit a golf ball with an initial velocity of 27 feet per second. The equation h = -6t^2 + 27t models the golf ball's height in feet t seconds after the ball was hit.

Explore the height of a golf ball hit with an initial velocity of 27 ft/s using the quadratic equation h = -6t^2 + 27t. Learn to find maximum height, time of flight, and height at specific times. Ideal for Grades 8-10.

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