Math Problem Statement

Question content area top 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?

Solution

To determine the time and maximum height of an object projected upward, we can analyze the height function s(t)=16t2+160ts(t) = -16t^2 + 160t, where:

  • tt is the time in seconds,
  • s(t)s(t) is the height in feet at time tt,
  • 16t2-16t^2 represents the effect of gravity in feet per second squared,
  • 160t160t 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)s(t). For a function of the form s(t)=at2+bt+cs(t) = at^2 + bt + c, the time tt at the vertex (or maximum point for a downward-opening parabola) is given by: t=b2at = -\frac{b}{2a} In our case:

  • a=16a = -16
  • b=160b = 160

Thus, t=1602(16)=16032=5 secondst = -\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=5t = 5 into the height function s(t)s(t) to find the maximum height: [ s(5) = -16(5)^2 +

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Math Problem Analysis

Mathematical Concepts

Quadratic Functions
Vertex of a Parabola
Projectile Motion

Formulas

Vertex formula for a quadratic function t = -b/(2a)
Height function s(t) = -16t^2 + 160t

Theorems

Vertex Theorem for Quadratic Functions

Suitable Grade Level

Grades 9-10