We are given the height equation of a baseball thrown 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 throw
• The astronaut is 6 ft 6 in. tall ($6.5$ ft)
• The initial velocity is 50 ft/sec
• Gravity on the moon causes an acceleration of $-2.7$ ft/sec²

Let's analyze the problem step by step.

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Part (a): Find the maximum height of the baseball.

The height equation follows the form of a quadratic function:

$s(t) = at^2 + bt + c$

where:

• $a = -2.7$,
• $b = 50$,
• $c = 6.5$.

The maximum height occurs at the vertex of the quadratic function. The time at which the ball reaches its maximum height is given by:

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

Substituting values:

$t = \frac{-50}{2(-2.7)}$

$t = \frac{50}{5.4}$

$t \approx 9.26 \text{ seconds}$

Now, we substitute $t = 9.26$ into $s(t)$ to find the maximum height:

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

$s(9.26) = -2.7(85.77) + 463 + 6.5$

$s(9.26) = -231.58 + 463 + 6.5$

$s(9.26) \approx 237.92 \text{ ft}$

So, the maximum height is about 237.92 ft.

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Part (b): When does the baseball hit the moon's surface?

The baseball hits the surface when $s(t) = 0$, so we solve:

$-2.7t^2 + 50t + 6.5 = 0$

Using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where:

• $a = -2.7$
• $b = 50$
• $c = 6.5$

First, calculate the discriminant:

$b^2 - 4ac = (50)^2 - 4(-2.7)(6.5)$

$= 2500 + 70.2$

$= 2570.2$

Now, take the square root:

$\sqrt{2570.2} \approx 50.7$

Now, solve for $t$:

$t = \frac{-50 \pm 50.7}{2(-2.7)}$

$t = \frac{-50 \pm 50.7}{-5.4}$

We get two solutions:

$t = \frac{-50 + 50.7}{-5.4} = \frac{0.7}{-5.4} \approx -0.13$

$t = \frac{-50 - 50.7}{-5.4} = \frac{-100.7}{-5.4} \approx 18.65$

Since time cannot be negative, we take $t \approx 18.65$ seconds.

So, the baseball hits the moon’s surface after about 18.65 seconds.

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Final Answers:

• Maximum height: 237.92 ft
• Time to hit the surface: 18.65 sec

Would you like more explanation or another approach?

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Related Questions:

1. What would the maximum height be if the initial velocity were increased to 60 ft/sec?
2. How long would it take to reach half of the maximum height?
3. How would the equation change if this were on Earth instead of the moon?
4. What is the velocity of the ball when it hits the ground?
5. How high would the ball go if thrown from a 10-ft-tall platform?

Tip:

For projectile motion problems, always identify the vertex for max height and solve $s(t) = 0$ for total time.