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To find the highest point that the object reaches, we need to determine the vertex of the quadratic function \( h(t) = -16t^2 + 160t + 1536 \). For a quadratic function in the form \( h(t) = at^2 + bt + c \), the time at which the maximum height occurs is given by:

\[
t = -\frac{b}{2a}
\]

Here, \( a = -16 \), \( b = 160 \), and \( c = 1536 \).

### Step 1: Find the time at which the maximum height occurs
\[
t = -\frac{160}{2(-16)} = \frac{160}{32} = 5 \text{ seconds}
\]

### Step 2: Calculate the height at \( t = 5 \)
Substitute \( t = 5 \) into the equation for \( h(t) \):

\[
h(5) = -16(5)^2 + 160(5) + 1536
\]
\[
h(5) = -16(25) + 800 + 1536
\]
\[
h(5) = -400 + 800 + 1536
\]
\[
h(5) = 1936 \text{ feet}
\]

Thus, the highest point that the object reaches is **1936 feet**.

Would you like more details or have any questions?

Here are some related questions:
1. How does changing the initial velocity affect the maximum height?
2. What is the significance of the coefficient of \( t^2 \) in the equation for vertical motion?
3. How long does it take for the object to hit the ground?
4. How would the equation change if the building was shorter or taller?
5. What would be the object's speed when it hits the ground?

**Tip:** The maximum or minimum of a quadratic function \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \), which can be used to find the vertex efficiently.

We are standing on the top of a 1536 feet tall building and launch a small object upward. The object's vertical position, measured in feet, after t seconds is h(t) = -16t^2 + 160t + 1536. What is the highest point that the object reaches?

Learn how to determine the maximum height of an object launched from a 1536-foot building using the quadratic equation h(t) = -16t^2 + 160t + 1536. This problem covers concepts such as the vertex formula for quadratic functions and parabolic motion. Suitable for students in Grades 9-12.

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