Problem 51 - Calculus Single Variable

Question

An object is thrown straight down. Its height at time \(t\) seconds is given in feet by \(H(t)=-16 t^{2}-160 t+84 .\) With what velocity does it impact the earth (at height 0).

Step-by-Step Solution

Answer

The object impacts the earth with a velocity of -176 feet per second.

Step 1: Find the Time of Impact

We need to find when the object hits the ground, at which point the height, \( H(t) \), is 0. Set the equation to zero and solve for \( t \): \[-16t^2 -160t + 84 = 0.\]First, divide the entire equation by -4 to simplify: \[4t^2 + 40t - 21 = 0.\]Use the quadratic formula, \[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\], where \(a = 4\), \(b = 40\), and \(c = -21\). Calculate the discriminant: \(b^2 - 4ac = 40^2 - 4(4)(-21) = 1600 + 336 = 1936.\)Find \( t \): \[t = \frac{-40 \pm \sqrt{1936}}{8}.\]\[t = \frac{-40 \pm 44}{8}.\]We get two solutions, \(t = \frac{4}{8} = 0.5\) and \(t = \frac{-84}{8} = -10.5\). The negative time does not make sense, so \(t = 0.5\) seconds.

Step 2: Differentiate to Find Velocity Function

To find the velocity at impact, differentiate the height function \( H(t) = -16t^2 -160t + 84 \) with respect to \( t \). This gives us the velocity function, \( V(t) \).Differentiate: \[V(t) = \frac{d}{dt}(-16t^2 -160t + 84) = -32t - 160.\]

Step 3: Calculate Velocity at Impact Time

Substitute \(t = 0.5\) seconds (time of impact) into the velocity function to find the speed of impact:\[V(0.5) = -32(0.5) - 160 = -16 - 160 = -176.\]The object impacts the earth with a velocity of \(-176\) feet per second. The negative sign indicates the direction is downward.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to solve quadratic equations of the form \( ax^2 + bx + c = 0 \). This formula provides the solutions for \( x \) as follows: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Here:

• \( a \), \( b \), and \( c \) are coefficients from the quadratic equation.
• The "\( \pm \)" symbol means there will be two solutions, one with plus and one with minus.

The formula is especially useful because it can provide solutions even when factoring is difficult. It's crucial for problems involving physics or real-life scenarios, like finding the time of impact for a falling object. In our object-in-motion example, using this formula allowed us to calculate the exact time when the object hits the ground, specifically solving the height function set to zero.

Derivative

The derivative of a function expresses how the function changes as its input changes. It's a fundamental concept in calculus used to determine the rate of change. For a function \( f(x) \), its derivative, denoted by \( f'(x) \) or \( \frac{df}{dx} \), provides the slope of the tangent line to the curve at any point.

• In physics, we use derivatives to find velocity, which is the rate of change of position with respect to time.
• The derivative tells us how quickly and in what manner the object's position changes over time.

In this problem, the height function \( H(t) = -16t^2 -160t + 84 \) is differentiated to find the velocity function. This allows us to evaluate the speed of the object at any given moment, especially at the impact time. Differentiation thus helps in transforming a position-time relationship into a velocity-time relationship.

Discriminant

The discriminant is a component within the quadratic formula, represented by \( b^2 - 4ac \), and it reveals the nature of the solutions for a quadratic equation.

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is exactly one real root (also called a repeated root or a double root).
• If the discriminant is negative, the roots are complex or imaginary.

In our exercise, the discriminant was calculated to be 1936, which is positive. This tells us there are two distinct times for which the object's height could be zero. However, considering the context, only the positive value of time makes physical sense, thus \( t = 0.5 \) second was the viable solution.

Differentiation

Differentiation is a calculus technique used for finding the derivative of a function. This process helps us understand how a function's output changes with respect to changes in its input.

• Differentiation is applied to derive acceleration, velocity, and other critical rates of change in various scientific fields.
• Through differentiation, we obtain a velocity function that describes an object's speed at any time \( t \).

In the process of solving our original exercise, we differentiate the height function \( H(t) \) to obtain the velocity function \( V(t) \). This gives us the formula \( V(t) = -32t - 160 \), which shows how the velocity of the object changes with time. This understanding allowed us to assess the exact rate of travel at the specific moment the object impacts the earth.