Problem 158 - Elementary Algebra

Question

An arrow is shot vertically upward at a rate of 220 feet per second. Use the projectile formula \(h=-16 t^{2}+v_{0} t\) to determine when height of the arrow will be 400 feet.

Step-by-Step Solution

Answer

The arrow will be at 400 feet after approximately 2.15 seconds.

Step 1 - Understand the Given Information

The formula we will use is the projectile formula: \[ h = -16t^2 + v_0 t \] where \(h = 400\) feet and \(v_0 = 220\) feet per second.

Step 2 - Set the Formula Equal to Height

Set the given height 400 feet equal to the projectile formula:\[ 400 = -16t^2 + 220t \]

Step 3 - Rearrange the Equation

Bring all terms to one side to set the equation to zero:\[ -16t^2 + 220t - 400 = 0 \]

Step 4 - Apply the Quadratic Formula

Use the quadratic formula to solve for \(t\): \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] with \(a = -16\), \(b = 220\), and \(c = -400\).

Step 5 - Calculate the Discriminant

Find the discriminant: \[ b^2 - 4ac = 220^2 - 4(-16)(-400) = 48400 - 25600 = 22800 \]

Step 6 - Find the Solutions for t

Substitute the discriminant and other values into the quadratic formula: \[ t = \frac{-220 \pm \sqrt{22800}}{2(-16)} \] \[ t = \frac{220 \pm 150.99}{-32} \] This gives us two solutions for \(t\): \[ t = \frac{220 + 150.99}{-32} \approx -11.58 \] and \[ t = \frac{220 - 150.99}{-32} \approx 2.15 \]

Step 7 - Choose the Positive Time

Since time cannot be negative, we discard the negative solution: \( t \approx 2.15 \) seconds.

Key Concepts

Quadratic Equations

Quadratic equations are essential in understanding projectile motion. They are second-degree polynomial equations of the form \( ax^2 + bx + c = 0 \). These equations can be solved using various methods, including factoring, completing the square, or the quadratic formula.

For our exercise, we use the quadratic formula, which is: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].

This formula helps us find the values for \( t \) when the equation cannot be easily factored. In our case, \( a = -16, b = 220, \text{and} c = -400 \). By substituting these values into the quadratic formula, we solve for \( t \), which represents the time when the arrow reaches the height of 400 feet.

Projectile Formula

The projectile formula is crucial to solving problems involving objects moving under the influence of gravity. The standard projectile formula is:
\[ h = -16 t^2 + v_0 t \]

where:

• \( h \) represents the height at time \( t \)
• \( -16t^2 \) accounts for the acceleration due to gravity in feet per second squared (assuming the object is shot vertically upward)
• \( v_0 \) is the initial velocity

For our problem, the formula helps us determine the height of an arrow shot at a rate of 220 feet per second. By setting \( h = 400 \) feet in the formula and rearranging terms, we derive the quadratic equation to solve for \( t \). This lets us calculate when the arrow reaches the specified height.

Solving for t

Solving for \( t \) involves using the quadratic formula. Here's a quick rundown of our specific problem:

• We first set the formula equal to the given height: \( 400 = -16t^2 + 220t \).
• Next, we rearrange all terms to one side: \( -16t^2 + 220t - 400 = 0 \).
• Then, we apply the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] with \( a = -16, b = 220, \) \text{and} \( c = -400 \).
• We calculate the discriminant: \[ b^2 - 4ac = 220^2 - 4(-16)(-400) = 48400 - 25600 = 22800 \]
• We find the two possible values of \( t \): \[ t = \frac{220 \pm 150.99}{-32} \]
• Finally, we discard the negative solution and choose the positive time: \( t \approx 2.15 \) seconds.

By following these steps, we determine that the arrow will reach the height of 400 feet approximately 2.15 seconds after being shot.