Problem 42 - Algebra and Trigonometry with Analytic Geometry

Question

An object is projected vertically upward with an initial velocity of \(v_{0} \mathrm{ft} / \mathrm{sec}\), and its distance \(s(t)\) in feet above the ground after \(t\) seconds is given by the formula \(s(t)=-16 t^{2}+v_{0} t\). (a) If the object hits the ground after 12 seconds, find its initial velocity \(v_{0}\). (b) Find its maximum distance above the ground.

Step-by-Step Solution

Answer

The initial velocity is 192 ft/sec and the maximum height is 576 feet.

Step 1: Understand the problem

We are given a formula for the distance of an object projected upward: \(s(t) = -16t^2 + v_0t\). We need to find the initial velocity \(v_0\) when the object hits the ground after 12 seconds and the maximum height it reaches.

Step 2: Set up the equation for the object hitting the ground

To find when the object hits the ground, we set \(s(t) = 0\). After 12 seconds, the equation is \(-16(12)^2 + v_0(12) = 0\). Simplify this equation.

Step 3: Solve for initial velocity \(v_0\)

The equation from Step 2 is \(-16 \times 144 + 12v_0 = 0\). Simplify it to \(-2304 + 12v_0 = 0\). Solving for \(v_0\) gives: \[ v_0 = \frac{2304}{12} = 192 \, \text{ft/sec} \].

Step 4: Find maximum height using vertex formula

The formula for height \(s(t)\) is quadratic in standard form \(ax^2 + bx + c = 0\). The vertex, giving the maximum height, occurs at \(t = -\frac{b}{2a}\). Here, \(a = -16\) and \(b = v_0 = 192\). Calculate \(t_{max} = \frac{192}{32} = 6\) seconds.

Step 5: Calculate maximum distance above the ground

Substitute \(t = 6\) back into the equation \(s(t) = -16t^2 + 192t\) to find \(s(6)\). This gives: \[ s(6) = -16 \times 36 + 192 \times 6 = -576 + 1152 = 576 \, \text{feet} \].

Key Concepts

Quadratic Equations

Projectile motion problems often involve quadratic equations, which are polynomial equations of the second degree. A quadratic equation typically takes the form:

• \(ax^2 + bx + c = 0\)

where \(a\), \(b\), and \(c\) are coefficients, and \(x\) is the variable. In our exercise, the equation \(s(t) = -16t^2 + v_0t\) describes the height of the object over time with coefficients \(a = -16\), \(b = v_0\), and \(c = 0\).
These equations often model real-world scenarios like the motion of objects under gravity. Solving quadratic equations can help find important values such as maximum height or when an object will hit the ground.
Several methods like factoring, using the quadratic formula, or completing the square can solve these equations.

Vertex Formula

The vertex formula is a key tool in analyzing the features of quadratic equations, specifically to find the maximum or minimum values, known as the vertex. For a quadratic equation \(ax^2 + bx + c\), the vertex can be found at:

• \(t = -\frac{b}{2a}\)

This formula helps in determining the time at which the maximum or minimum value occurs in projectile motion. In our problem, with \(a = -16\) and \(b = 192\), substituting these values gives us \(t_{max} = \frac{192}{32} = 6\) seconds.
This point represents when the object reaches its highest distance above the ground, where the velocity changes direction due to gravity.

Initial Velocity

Initial velocity, often denoted as \(v_0\), is critical in projectile problems as it defines how fast and in what direction an object begins its motion. In our exercise, the initial velocity can be determined by using the condition set when the object hits the ground. The ground contact condition in this scenario is represented by setting the equation \(s(t) = 0\) after 12 seconds.
Using the equation

• \(-16(12)^2 + v_0(12) = 0\)

we simplify to solve for \(v_0 = \frac{2304}{12} = 192 \text{ ft/sec}\).
Knowing the initial velocity is essential for predicting how far the object will travel or what maximum height it will reach.

Maximum Height

The maximum height of a projectile in motion gives us the peak point of its path. To find this, we use the height equation and the vertex formula. This procedure involves firstly, determining the time at which the maximum height is reached using:

• \(t = -\frac{b}{2a}\)

and secondly, substituting this time back into the height function.
For our exercise, we already calculated \(t_{max} = 6\) seconds. By substituting back into the height function

• \(s(t) = -16t^2 + 192t\)

we find \(s(6) = -16 \times 36 + 192 \times 6 = 576 \text{ feet}\).
This point on the path represents the highest point reached by the object before gravity pulls it back down.