Problem 50 - Calculus Single Variable

Question

Concern an object that is propelled straight up. Its height at time \(t\) seconds is given in feet by \(H(t)=-16 t^{2}+128 t+68\). Supposing that the trajectory is such that the object is able to fall to the ground (height 0 feet), for how many seconds does the object fall?

Step-by-Step Solution

Answer

The object falls for 4.5 seconds after reaching its highest point.

Step 1: Identify the Objective

We need to find out the time duration for which the object falls back to the ground from its highest point.

Step 2: Determine the Time to Hit the Ground

Set the height equation to zero to solve for the time when the object hits the ground: \[ H(t) = -16t^2 + 128t + 68 = 0 \]

Step 3: Apply the Quadratic Formula

Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = -16 \), \( b = 128 \), and \( c = 68 \) to find the two solutions for \( t \).

Step 4: Calculate the Discriminant

Evaluate the discriminant \( b^2 - 4ac \): \[ 128^2 - 4(-16)(68) = 16384 + 4352 = 20736 \]

Step 5: Solve for Time Using the Quadratic Formula

Substitute back into the quadratic formula to solve for \( t \): \[ t = \frac{-128 \pm \sqrt{20736}}{-32} \]This simplifies to \[ t = \frac{-128 \pm 144}{-32} \]

Step 6: Calculate Potential Times

Solve for \( t \):For \( t_1 = \frac{-128 + 144}{-32} = -0.5 \) is not physically meaningful, so Use \( t_2 = \frac{-128 - 144}{-32} = 8.5 \) seconds.

Step 7: Determine the Duration of the Fall

Subtract the time at the highest point from the total time the object is in the air. First, find the highest point: midway between the roots of the equation, \( t = \frac{-b}{2a} = 4 \). This means it took 4 seconds to reach its peak, so it takes 4.5 seconds to fall to the ground.

Key Concepts

Physics Applications

When it comes to understanding quadratic equations, physics provides a fascinating context, particularly when dealing with projectile motion. In such scenarios, the motion of objects through space can often be described using quadratic equations. This is especially true when an object moves under the influence of gravitational forces, with air resistance typically being ignored for simplicity.

In our exercise, we observe a scenario where an object is propelled straight upwards. The equation provided, \( H(t) = -16t^2 + 128t + 68 \), models the height of this object at any given time \( t \). The coefficients in this equation hold specific physical significance.

• The term \(-16t^2\) is influenced by gravity, specifically representing how gravity decelerates the object as it ascends and accelerates it when descending, with \(-16\) reflecting the constant acceleration due to gravity in feet per second squared.
• The term \(128t\) represents the initial velocity of the object, showing how its velocity contributes positively to its height initially.
• Finally, \(68\) gives us the initial height from which the object was launched.

Understanding these components helps us appreciate how physics applications simplify real-world problems into mathematically solvable equations.

Projectile Motion

Projectile motion, as a concept, is indispensable to understanding how objects behave in space when launched or thrown. This type of motion is characterized by the object following a curved trajectory called a parabola.

In this particular exercise, the object follows a vertical path. This is a special case of projectile motion where horizontal components are absent. The object rises until the force of gravity entirely outweighs its initial push upwards, and then it falls back down, landing where it started or at another level of interest, such as the ground, which is our case.

The height-time function, \( H(t) = -16t^2 + 128t + 68 \), maps the object's journey. Initially, the object moves upward, where its velocity decreases due to gravity, reaches a peak, and then accelerates downwards.

• Understanding the time it takes to reach each part of its journey is crucial. The peak is its highest point, found using \( t = \frac{-b}{2a} \).
• From this peak, it falls back to the ground as indicated by setting \( H(t) = 0 \).

By looking at the projectile motion in this way, students gain insight into how objects behave when subjected to initial forces and then left to the whims of gravity.

Quadratic Formula

In solving problems involving quadratic equations, the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is an invaluable tool. It helps us find the zeros or roots of the equation, which are the solutions we're looking for.

For our exercise, applying this formula allows us to find when the object will return to the ground. This is done by setting the height equation \( H(t) \) to zero and solving for \( t \).

• First, determine the coefficients where \( a = -16 \), \( b = 128 \), and \( c = 68 \) in the equation \( H(t) = -16t^2 + 128t + 68 \).
• Plug these values into the quadratic formula to solve for \( t \): \( t = \frac{-128 \pm \sqrt{20736}}{-32} \).
• The two possible solutions for \( t \) indicate potential times at which the object could be at ground level. However, the feasible time is the positive one, \( t = 8.5 \) seconds, because negative time does not make sense in this context.

Applying the quadratic formula simplifies identifying the projectile's behavior over time and helps illustrate the object's full journey.