ALGEBRA l-B-1200380B2/205 $$ 7-3 / 7-4 $$ : Lesson Quiz 24-25 A ball is thrown upward with an initial velocity of 75 feet per second and an initial height of 4 feet. Given $$ h(t)=-16 t^{2}+v_{0} t+h_{0} $$, complete function $$ h $$ to model the vertical motion of the ball. Then find the ball's maximum height, to the nearest foot. $$ h(t)=-16 t^{2}+\square t+\square $$

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We start with the general formula for the height of the ball:  
$$
h(t) = -16t^2 + v_0t + h_0.
$$

Given that $$ v_0 = 75 $$ feet per second and $$ h_0 = 4 $$ feet, we substitute these values into the equation:  
$$
h(t) = -16t^2 + 75t + 4.
$$

The ball will reach its maximum height at the time  
$$
t = -\frac{b}{2a},
$$
where $$ a = -16 $$ and $$ b = 75 $$. Thus,  
$$
t = -\frac{75}{2(-16)} = \frac{75}{32}.
$$

Next, we find the maximum height by substituting $$ t = \frac{75}{32} $$ into the height function:
$$
h\left(\frac{75}{32}\right) = -16 \left(\frac{75}{32}\right)^2 + 75\left(\frac{75}{32}\right) + 4.
$$

Step 1: Compute $$\left(\frac{75}{32}\right)^2$$:
$$
\left(\frac{75}{32}\right)^2 = \frac{75^2}{32^2} = \frac{5625}{1024}.
$$

Step 2: Multiply by $$-16$$:
$$
-16 \times \frac{5625}{1024} = -\frac{16 \times 5625}{1024} = -\frac{90000}{1024}.
$$

Step 3: Compute the second term:
$$
75\left(\frac{75}{32}\right) = \frac{75^2}{32} = \frac{5625}{32}.
$$

Step 4: Now, express the function:
$$
h\left(\frac{75}{32}\right) = -\frac{90000}{1024} + \frac{5625}{32} + 4.
$$

To combine the terms, we convert $$\frac{5625}{32}$$ to have a common denominator of 1024:
$$
\frac{5625}{32} = \frac{5625 \times 32}{32 \times 32} = \frac{180000}{1024}.
$$

Similarly, write $$4$$ as:
$$
4 = \frac{4 \times 1024}{1024} = \frac{4096}{1024}.
$$

Now, combine all terms:
$$
h\left(\frac{75}{32}\right) = -\frac{90000}{1024} + \frac{180000}{1024} + \frac{4096}{1024}.
$$

Combine the numerators:
$$
-\;90000 + 180000 + 4096 = 94096.
$$

Thus,
$$
h\left(\frac{75}{32}\right) = \frac{94096}{1024}.
$$

Divide to simplify:
$$
\frac{94096}{1024} \approx 91.89 \text{ feet}.
$$

Rounding to the nearest foot, the ball's maximum height is approximately  
$$
92 \text{ feet}.
$$

The completed function and maximum height are:

$$
h(t)=-16t^2+75t+4 \quad \text{and} \quad \text{Maximum height} \approx 92 \text{ feet}.
$$
The ball's height is modeled by $$ h(t) = -16t^2 + 75t + 4 $$. The maximum height reached by the ball is approximately 92 feet.

ALGEBRA l-B-1200380B2/205
: Lesson Quiz 24-25
A ball is thrown upward with an initial velocity of 75 feet per second and an initial height of 4 feet. Given , complete function to model the vertical motion of the ball. Then find
the ball’s maximum height, to the nearest foot.

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ALGEBRA l-B-1200380B2/205 : Lesson Quiz 24-25 A ball is thrown upward with an initial velocity of 75 feet per second and an initial height of 4 feet. Given , complete function to model the vertical motion of the ball. Then find the ball’s maximum height, to the nearest foot.