Problem 41 - Precalculus Student Solutions Manual 5th

Question

A projectile is launched from ground level with an initial velocity of \(v_{0}\) feet per second. Neglecting air resistance, its height in feet \(t\) seconds after launch is given by $$s=-16 t^{2}+v_{0} t$$ Find the time(s) that the projectile will (a) reach a height of 80 \(\mathrm{ft}\). and (b) return to the ground for the given value of \(v_{0}\). Round answers to the nearest hundredth if necessary. $$v_{0}=32$$

Step-by-Step Solution

Answer

The projectile returns to the ground at t = 2 seconds.

Step 1: Set Up the Equation for Reaching 80 feet

To find the time when the projectile reaches a height of 80 feet, set the height equation equal to 80: \[ -16t^2 + 32t = 80 \]

Step 2: Rearrange the Equation

Subtract 80 from both sides to set the equation to 0: \[ -16t^2 + 32t - 80 = 0 \]

Step 3: Solve the Quadratic Equation

Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \(a = -16\), \(b = 32\), and \(c = -80\). Calculate the discriminant first: \[ b^2 - 4ac = 32^2 - 4(-16)(-80) = 1024 - 5120 = -4096 \]. Since the discriminant is negative, the answer has no real solution.

Step 4: Short Answer for Part (a)

There are no real times at which the projectile reaches 80 feet.

Step 4: Set Up the Equation for Returning to Ground

To find the time when the projectile returns to the ground, set the height equation equal to 0: \[ -16t^2 + 32t = 0 \]

Step 5: Factor the Quadratic Equation

Factor out \(t\) from the equation: \[ t(-16t + 32) = 0 \]

Step 6: Solve for t

Set each factor equal to 0 and solve: \[ t = 0 \] and \[ -16t + 32 = 0 \implies t = 2 \]

Key Concepts

Quadratic Equations

A quadratic equation is a polynomial equation of degree 2, generally written as \(ax^2 + bx + c = 0\). These equations have various methods of solving them:

• Factoring: Expressing the quadratic in a product of simpler expressions and then setting each equal to zero.
• Quadratic Formula: An equation \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) that gives solutions directly. It is useful when factoring is difficult.
• Completing the Square: Rewriting the quadratic in the form \((x+d)^2 = e\) and solving for \(x\) by taking the square root of both sides.

In this particular exercise, we used the quadratic formula, yet we found that the discriminant \(b^2 - 4ac\) was negative, indicating no real solution exists for the quadratic equation in some cases.

Height Equation

In projectile motion, the height equation is a quadratic equation that describes how the height of a projectile changes over time. The height equation has the form:
\ s = -16t^2 + v_0 t \
Here:

• \(s\) denotes the height at time \(t\)
• \(t\) represents time in seconds after the projectile is launched
• The term \(-16t^2\) corresponds to the effect of gravity, with \(-16\) being half of the gravity constant in feet per second squared (since gravitational acceleration is approximately \(32 \frac{ft}{s^2}\))
• \(v_0\) represents the initial velocity of the projectile in feet per second

The height equation captures the parabolic trajectory that any projectile follows under the effects of gravity alone.

Initial Velocity

Initial velocity, denoted as \(v_0\), is a crucial parameter in projectile motion equations. This represents the speed at which a projectile is launched, measured in feet per second (or other units of speed). Initial velocity affects:

• Maximum Height: The highest point the projectile reaches. Higher initial velocity means a higher peak.
• Range: The horizontal distance covered. More initial velocity results in a farther range.
• Time of Flight: The total time the projectile remains in the air. Greater initial velocity typically extends the flight time.

In the given exercise, the initial velocity \(v_0\) is \(32 \frac{ft}{s}\). This value is used to determine the height of the projectile at any given time \(t\) after it is launched. With the height equation \(-16t^2 + 32t\), we analyze how long it takes for the projectile to return to the ground and explore other significant moments such as reaching maximum height.