Projectile Motion: Finding Flight Time for a Tossed Ball

What You Will Learn

• Projectile motion modeling — How to set up the standard quadratic height equation for objects under gravity
• Physical parameter identification — Distinguishing between initial height, launch velocity, and target height in word problems
• Quadratic solution interpretation — Why projectile problems yield two solutions and which one represents the physical answer
• Precision and rounding — Working with irrational roots and applying appropriate significant figures to real-world contexts
• Verification strategies — How to check projectile motion answers by substitution and reasonableness testing

Solution: Method 1 — Standard Projectile Motion Equation

For any object moving under constant gravitational acceleration, the height at time t follows a quadratic pattern. We'll use the standard physics formula and solve for when the ball reaches the catch height.

Step 1 — Set up the height equation

The height of a projectile at time t seconds is given by:

Where g = 32 ft/s² (gravitational acceleration), v₀ = 25 ft/s (initial velocity), and h₀ = 4 ft (starting height).

Step 2 — Substitute the known values

Plugging in our values:

Step 3 — Set height equal to catch position

We want to find when h(t) = 5 feet (the catch height):

Step 4 — Rearrange to standard quadratic form

Moving all terms to one side:

Step 5 — Apply the quadratic formula

Using t = (-b ± √(b² - 4ac)) / 2a with a = 16, b = -25, c = 1:

Step 6 — Calculate the discriminant

Computing √561 ≈ 23.686:

This gives us two solutions:

• t₁ = (25 + 23.686) / 32 = 48.686 / 32 ≈ 1.521
• t₂ = (25 - 23.686) / 32 = 1.314 / 32 ≈ 0.041

Step 7 — Choose the physical solution

The smaller time (0.041 s) represents when the ball passes 5 feet on the way up. The larger time (1.521 s) is when it passes 5 feet on the way down — this is when the juggler catches it.

Rounded to the nearest tenth: 1.5 seconds

Solution: Method 2 — Vertex Form & Symmetry Analysis

This approach uses the symmetry of parabolic motion. We find when the ball reaches its maximum height, then use the parabola's properties to locate the catch time.

Step 1 — Find the vertex time

The ball reaches maximum height when the velocity is zero. From h(t) = -16t² + 25t + 4, the vertex occurs at:

Step 2 — Calculate maximum height

Substituting back:

Step 3 — Find when ball drops to 5 feet

Since the parabola is symmetric around the vertex, we need to find how long after the vertex the ball takes to drop from 13.765 ft to 5 ft. The ball drops:

Step 4 — Use kinematic equation for the downward journey

Starting from the vertex with zero initial velocity, the time to fall 8.765 feet is:

Step 5 — Add vertex time and fall time

Verification

Let's confirm our answer by substituting t = 1.5 back into the original height equation:

This is very close to our target of 5 feet. The small difference (0.5 ft) comes from rounding 1.521 to 1.5. Using the exact value 1.521 seconds:

Watch Out For These

The Math Beneath the Surface

This problem exemplifies the general principle that projectile motion under constant gravity always produces a quadratic height function. The standard form is:

Every projectile motion problem reduces to this equation, regardless of whether it involves balls, bullets, or bridges. The key insight is that any height along the trajectory is crossed twice — once ascending, once descending — which is why quadratic equations in physics typically yield two solutions.

Reality Check

Does 1.5 seconds seem reasonable for a juggler's toss? Let's think about this:

Maximum height check: In our solution, the ball reaches about 13.8 feet high — that's reasonable for a juggler throwing upward with significant force.

Speed check: An initial velocity of 25 ft/s equals about 17 mph — fast but not impossibly so for a skilled juggler's upward throw.

Timing check: 1.5 seconds total flight time means about 0.75 seconds up and 0.75 seconds down. Try counting "one-mississippi" — that's about the right duration for watching a ball arc overhead.

All of these align with what we'd expect from real juggling, confirming our mathematical answer makes physical sense.