We are given the height function for a baseball thrown by an astronaut on the moon:

$s(t) = -2.7t^2 + 50t + 6.5$

where:

• $s(t)$ is the height of the baseball in feet,
• $t$ is the time in seconds after the ball was thrown,
• The initial height is $6.5$ feet (since the astronaut is 6 feet 6 inches tall),
• The initial velocity is 50 feet per second.

Part A: Find the time when the baseball reaches its maximum height.

The function $s(t) = -2.7t^2 + 50t + 6.5$ is a quadratic equation. Since the coefficient of $t^2$ is negative, the parabola opens downward, meaning the maximum height is at the vertex of the parabola. The time at which this maximum occurs is given by the formula:

$t = \frac{-b}{2a}$

For our equation, $a = -2.7$ and $b = 50$, so:

$t = \frac{-50}{2(-2.7)} = \frac{-50}{-5.4} = 9.26 \, \text{seconds}$

Thus, the ball reaches its maximum height at $t = 9.26$ seconds.

Part B: Find the maximum height reached by the baseball.

To find the maximum height, substitute $t = 9.26$ into the height equation $s(t)$:

$s(9.26) = -2.7(9.26)^2 + 50(9.26) + 6.5$

Now, let’s calculate this.

$s(9.26) = -2.7(85.77) + 50(9.26) + 6.5$ $s(9.26) = -231.579 + 463 + 6.5$ $s(9.26) = 237.921 \, \text{feet}$

Thus, the maximum height reached by the baseball is approximately $237.92$ feet.

Summary:

• The ball reaches its maximum height at $t = 9.26$ seconds.
• The maximum height is approximately $237.92$ feet.

Let me know if you'd like further details or clarification!

---

Here are 5 related questions:

1. What is the height of the baseball after 5 seconds?
2. How long does it take for the ball to hit the ground?
3. What is the velocity of the ball at its maximum height?
4. What would the maximum height be if the initial velocity were increased to 60 ft/sec?
5. How would the graph of the height function change if the ball were thrown with a different initial velocity?

Tip: The vertex of a quadratic function is key to finding the maximum or minimum value in motion problems like this one.