Problem 78 - Brief Applied Calculus

Question

GENERAL: Impact Time of a Projectile If an object is thrown upward so that its height (in feet) above the ground \(t\) seconds after it is thrown is given by the function \(h(t)\) below, find when the object hits the ground. That is, find the positive value of \(t\) such that \(h(t)=0\). Give the answer correct to two decimal places. [Hint: Enter the function in terms of \(x\) rather than t. Use the ZERO operation, or TRACE and ZOOM IN, or similar operations.] $$ h(t)=-16 t^{2}+40 t+4 $$

Step-by-Step Solution

Answer

The object hits the ground at approximately \( t = 2.59 \) seconds.

Step 1: Set up the Equation

The object hits the ground when its height is zero, so we need to find the value of \( t \) for which \( h(t) = 0 \). Substitute into the equation: \(-16t^2 + 40t + 4 = 0\).

Step 2: Solve Using the Quadratic Formula

To solve \(-16t^2 + 40t + 4 = 0\), use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), where \( a = -16 \), \( b = 40 \), and \( c = 4 \).

Step 3: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac = 40^2 - 4(-16)(4) = 1600 + 256 = 1856 \).

Step 4: Find the Roots

Substitute the values into the quadratic formula: \( t = \frac{-40 \pm \sqrt{1856}}{-32} \). Calculate \( \sqrt{1856} \approx 43.07 \).

Step 5: Evaluate the Positive Root

Find the positive value of \( t \): \( t = \frac{40 + 43.07}{32} \approx 2.59 \).

Step 6: Verification

Verify using a calculator or graphing tool to ensure \( h(t) = 0 \) at \( t \approx 2.59 \) and confirm it is the correct, positive root.

Key Concepts

Projectile Motion

Projectile motion is the path an object follows when it is launched into the air and influenced by gravity. Understanding this concept is crucial when dealing with problems involving objects thrown upwards or at an angle. Typically, the motion can be broken down into two components: horizontal and vertical. In this exercise, the problem focuses on the vertical component of the motion where the height of the object is modeled by a quadratic equation.

• The vertical motion is dictated by gravity, impacting the object’s speed and height over time.
• The initial velocity and position determine the starting point of the projectile’s path. In this case, the initial velocity is 40 ft/s, and it starts 4 feet above the ground.
• The equation for the object's height is given as a quadratic function of time: \( h(t)=-16t^2+40t+4 \).

The coefficients in the equation indicate the influence of gravity, initial velocity, and initial position. Solving this equation helps us determine when the object will hit the ground, which is one practical application of projectile motion.

Discriminant

The discriminant is a specific part of the quadratic formula that provides important information about the nature of the roots. It is calculated from the expression under the square root sign in the quadratic formula: \( b^2 - 4ac \).

• A positive discriminant indicates that there are two distinct real roots.
• If the discriminant is zero, the quadratic equation has exactly one real root, meaning the object will touch the ground at exactly one point.
• A negative discriminant would imply that there are no real roots, which does not apply to projectile motion problems as these involve real and practical solutions.

In this exercise, the discriminant calculated was \( 1856 \). Since it is positive and greater than zero, it assures us that the object will have two moments in time where it can meet the height of zero. However, since projectile motion occurs over time in one direction, we only consider the positive time value.

Roots of a Function

The roots of a function are values of the variable that make the function equal to zero. In the context of projectile motion, finding the roots of the height function tells us the time instances when the object hits the ground. Here’s how we handle it using the quadratic equation:

• Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find \( t \).
• Substitute \( a = -16 \), \( b = 40 \), and \( c = 4 \) into the formula.
• Calculate the root, ensuring to solve for the only relevant root, which is the positive time when the object hits the ground.

In this problem, solving with these parameters yields two time values, but only the positive root is physically meaningful. For this scenario, we calculated \( t \approx 2.59 \) seconds, which is when the object hits the ground. Understanding the roots of the function in projectile motion is key since it directly correlates to vital events in the trajectory of the projectile.