Projectile Motion: Solving Quadratic Equations for Flight Time

What This Problem Teaches

• Setting up and solving quadratic equations from physics formulas
• Using the quadratic formula when factoring isn't practical
• Interpreting negative solutions in real-world contexts
• Understanding the physical meaning of projectile motion equations
• Converting between mathematical solutions and meaningful answers

Visualizing the Problem

The diagram shows the ball's trajectory from launch to impact. We need to find when the height equals zero.

Solution: Method 1 — The Quadratic Formula Approach

When the ball hits the ground, its height is zero. We need to solve the equation when h(t) = 0.

Step 1 — Set up the equation

The ball hits the ground when h(t) = 0, so we solve:

Step 2 — Identify the quadratic coefficients

In the standard form at² + bt + c = 0, we have:

• a = -16
• b = 147
• c = 6.5

Step 3 — Apply the quadratic formula

Using t = (-b ± √(b² - 4ac)) / (2a):

Step 4 — Calculate the discriminant

First, let's compute b² - 4ac:

Since √22,025 = 148.41 (approximately), we get:

Step 5 — Find both solutions

This gives us two solutions:

• t = (-147 + 148.41) / (-32) = 1.41 / (-32) ≈ -0.044
• t = (-147 - 148.41) / (-32) = -295.41 / (-32) ≈ 9.23

Step 6 — Choose the physically meaningful solution

Since time cannot be negative in this context, we take the positive solution: t ≈ 9.23 seconds.

Solution: Method 2 — Completing the Square Approach

We can also solve by completing the square, which reveals insights about the maximum height along the way.

Step 1 — Factor out the coefficient of t²

Starting with -16t² + 147t + 6.5 = 0, factor out -16:

Step 2 — Complete the square inside the parentheses

To complete the square for t² - 9.1875t, we add and subtract (9.1875/2)² = (4.59375)² ≈ 21.10:

Step 3 — Simplify and solve

Expanding:

Step 4 — Take the square root

Taking the square root of both sides:

This gives us t ≈ 9.23 or t ≈ -0.05, confirming our previous result.

Verification

Let's verify our answer by substituting t = 9.23 back into the original equation:

The result is very close to zero (0.27 feet), confirming our answer. The small difference is due to rounding in our calculations.

We can also verify that the negative solution doesn't make physical sense. At t = -0.044:

While mathematically correct, negative time represents a moment before the ball was thrown, which isn't physically meaningful for our problem.

Watch Out For These Pitfalls

The Physics Behind the Formula

The height equation h(t) = -16t² + 147t + 6.5 comes from the kinematic equation for motion under constant acceleration:

Where:

• h₀ = 6.5 feet (initial height)
• v₀ = 147 ft/s (initial velocity)
• a = -32 ft/s² (acceleration due to gravity)
• ½a = ½(-32) = -16

This formula applies to any projectile in Earth's gravitational field, assuming air resistance is negligible. The -16t² term dominates for large values of t, which is why the ball eventually comes down despite its initial upward velocity.

Real Applications

Sports ballistics: Baseball analysts use similar equations to predict where fly balls will land, helping outfielders position themselves optimally.

Engineering safety: When designing protective barriers around construction sites, engineers calculate projectile paths to determine safe distances from falling objects.

Military and aerospace: Artillery targeting and rocket trajectory calculations use these same quadratic motion principles, though with additional complexity for air resistance and rotation effects.