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Part 1
If an object is projected upward from ground level with an initial velocity of 160 ft per​ sec, then its height in feet after t seconds is given by ​s(t)equalsminus16tsquaredplus160t. Find the number of seconds it will take to reach its maximum height. What is this maximum​ height?


To determine the time and maximum height of an object projected upward, we can analyze the height function \( s(t) = -16t^2 + 160t \), where:

- \( t \) is the time in seconds,
- \( s(t) \) is the height in feet at time \( t \),
- \( -16t^2 \) represents the effect of gravity in feet per second squared,
- \( 160t \) represents the initial velocity of 160 feet per second.

### Step 1: Determine the time to reach the maximum height
The maximum height of a projectile occurs at the vertex of the parabolic function \( s(t) \). For a function of the form \( s(t) = at^2 + bt + c \), the time \( t \) at the vertex (or maximum point for a downward-opening parabola) is given by:
\[
t = -\frac{b}{2a}
\]
In our case:
- \( a = -16 \)
- \( b = 160 \)

Thus,
\[
t = -\frac{160}{2 \cdot (-16)} = \frac{160}{32} = 5 \text{ seconds}
\]

So, it takes 5 seconds to reach the maximum height.

### Step 2: Calculate the maximum height
Now, substitute \( t = 5 \) into the height function \( s(t) \) to find the maximum height:
\[
s(5) = -16(5)^2 +

If an object is projected upward from ground level with an initial velocity of 160 ft per sec, then its height in feet after t seconds is given by s(t) = -16t^2 + 160t. Find the number of seconds it will take to reach its maximum height. What is this maximum height?

Learn how to calculate the time and maximum height of a projectile launched with an initial velocity of 160 feet per second using the quadratic function s(t) = -16t^2 + 160t. This problem covers core concepts of quadratic functions, vertex formula application, and projectile motion, ideal for students in Grades 9-10.

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