Rocket Projectile Motion: Height Above Launch Pad

What You Will Learn

• Finding the vertex of a parabola to determine maximum values in quadratic motion
• Interpreting the physical meaning of each term in a projectile motion equation
• Solving quadratic equations by factoring to find when objects reach ground level
• Distinguishing between mathematical and physical solutions in real-world contexts
• Connecting algebraic techniques to physics concepts like gravity and initial velocity

Visualizing the Rocket's Path

Solution: Method 1 — The Vertex Formula Approach

Since the rocket's height follows a quadratic equation, its path forms a parabola. For any parabola y = ax² + bx + c, the maximum (or minimum) occurs at the vertex. Let's use this to find both answers systematically.

Step 1 — Find when maximum height occurs

For the quadratic h(t) = -5t² + 40t + 45, we have a = -5 and b = 40. The vertex occurs at:

Step 2 — Calculate the maximum height

Substitute t = 4 back into the height equation:

Step 3 — Find the flight time by setting height to zero

The rocket lands when its height equals zero. Set the equation equal to zero and solve:

Step 4 — Factor the equation

First, factor out -5 to simplify:

Now factor the quadratic expression. We need two numbers that multiply to -9 and add to -8:

Step 5 — Solve for time values

This gives us t = 9 or t = -1. Since negative time doesn't make physical sense (the rocket launches at t = 0), the flight time is 9 seconds.

Solution: Method 2 — Physics-Based Derivative Approach

We can also solve this using calculus, which connects directly to the physics of motion. The derivative of position gives velocity, and at the highest point, velocity equals zero.

Step 1 — Find the velocity function

Take the derivative of the height function:

Step 2 — Set velocity to zero for maximum height

At the peak, the rocket momentarily stops before falling:

Step 3 — Calculate maximum height

Substitute t = 4 into the original height equation:

Step 4 — Find landing time using the quadratic formula

For -5t² + 40t + 45 = 0, using a = -5, b = 40, c = 45:

This gives t = (-40 + 50)/(-10) = -1 or t = (-40 - 50)/(-10) = 9. Taking the positive solution, the flight time is 9 seconds.

Verification

Let's verify both answers by checking our work:

Checking maximum height

At t = 4 seconds:

Checking flight time

At t = 9 seconds (landing):

Physical reasonableness check

The rocket starts at 45m, goes up to 125m (gaining 80m), then falls back down. With an initial velocity of 40 m/s upward, this height gain and 9-second total flight time align with what we'd expect under Earth's gravity.

Common Pitfalls

The Physics Behind the Math

This problem demonstrates the standard form of projectile motion under constant acceleration. The general equation is:

Where:

• h₀ = 45m is the initial height
• v₀ = 40 m/s is the initial velocity
• g = 10 m/s² is gravitational acceleration
• ½g = 5, which explains the -5 coefficient

This gives us h(t) = 45 + 40t - 5t², which matches our equation when rearranged. Every quadratic motion problem follows this same pattern—the maximum occurs at t = v₀/g, and the object returns to its starting height at t = 2v₀/g.

Real Applications

• Aerospace engineering: Mission planners use these calculations to determine fuel requirements and trajectory windows for rocket launches
• Sports science: Analyzing the trajectory of basketballs, footballs, or javelins to optimize athletic performance
• Safety engineering: Calculating safe distances for fireworks displays and determining where debris might land
• Military ballistics: Predicting artillery shell trajectories and determining optimal firing angles for maximum range