Problem 49 - Beginning and Intermediate Algebra

Question

If an object on Earth is projected upward with an initial velocity of \(32 \mathrm{ft}\) per sec, then its height after \(t\) seconds is given by $$s(t)=-16 t^{2}+32 t$$Find the maximum height attained by the object and the number of seconds it takes to hit the ground.

Step-by-Step Solution

Answer

The maximum height is 16 feet, and the object hits the ground after 2 seconds.

Step 1: Identify the function

The height of the object as a function of time is given by the equation \(s(t) = -16t^2 + 32t\). This is a quadratic function.

Step 2: Find the time to reach maximum height

For a quadratic equation of the form \(ax^2 + bx + c\), the maximum (or minimum) value is found at \(t = -\frac{b}{2a}\). Here, \(a = -16\) and \(b = 32\). Plug these values into the formula: \(t = -\frac{32}{2(-16)} = 1\).

Step 3: Calculate the maximum height

To find the maximum height, substitute \(t = 1\) back into the height function \(s(t) = -16t^2 + 32t\): \(s(1) = -16(1)^2 + 32(1) = -16 + 32 = 16\). So, the maximum height attained by the object is \(16\) feet.

Step 4: Find the time when the object hits the ground

Set the height function \(s(t)\) equal to \(0\) to find when the object hits the ground: \(0 = -16t^2 + 32t\). Solve by factoring: \(0 = t(-16t + 32)\). This gives two solutions: \(t = 0\) and \(t = 2\). The object hits the ground after \(2\) seconds.

Key Concepts

Maximum height

In projectile motion, the maximum height is the peak point that an object reaches before it starts descending. For quadratic functions describing projectile motion, the formula for maximum height can be found using the vertex formula. The equation given was: \[s(t) = -16t^2 + 32t\]. Here, the coefficients are \( a = -16 \) and \( b = 32 \). We use the vertex formula \( t = -\frac{b}{2a} \). Plugging in the values, we get \[ t = -\frac{32}{2 * -16} = 1 \]. So, it takes 1 second to reach the maximum height. To find the maximum height itself, substitute \( t = 1 \) back into the height equation: \[ s(1) = -16(1)^2 + 32(1) = 16 \]. Hence, the maximum height attained by the object is 16 feet.

Time of flight

The time of flight refers to the total duration that an object remains in the air. It starts from the moment the object is launched until it touches the ground again. For the given quadratic function \(s(t) = -16t^2 + 32t\), the object hits the ground when its height \(s(t)\) becomes zero. Setting the equation to zero for this calculation: \[ 0 = -16t^2 + 32t \] We factorize this to get: \[ 0 = t(-16t + 32) \]. Which gives two solutions: \[ t = 0 \] and \[ t = 2 \]. The solution \( t = 0 \) refers to the initial time of projection, and \( t = 2 \) indicates the time the object hits the ground. Thus, the total time of flight is 2 seconds.

Projectile motion

Projectile motion describes the trajectory of an object launched into the air and influenced only by gravity. In the absence of air resistance, the motion follows a parabolic path. The given function \( s(t) = -16t^2 + 32t \) represents the height of an object projected upwards. This form - a quadratic equation - is typical for projectile motion. Key aspects to focus on:

• Initial velocity: Here, it is 32 feet per second.
• Acceleration due to gravity: Represented by the coefficient -16.
• Path: The parabolic path traced by the object.

Understanding these points helps in analyzing various aspects like maximum height and time of flight.

Solving quadratic equations

To solve quadratic equations, especially in the context of projectile motion, several methods can be used. For our equation \(s(t) = -16t^2 + 32t\), these are common approaches:

• Factoring: Set the equation equal to zero and solve for \(t\). Example: \(0 = t(-16t + 32)\).
• Quadratic Formula: For general form \(ax^2 + bx + c = 0\), use \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This works when factoring is complex.
• Completing the Square: Another method, though often less straightforward, involves creating a perfect square trinomial from the quadratic equation.

In the given exercise, factoring was the fastest method to find the time the object hits the ground. Choose the method based on the specific form and complexity of the quadratic equation.