Projectile Motion: Solving for Time at Given Height
What This Problem Teaches
- How to set up and solve quadratic equations from physics formulas
- Understanding that projectile problems typically have two time solutions
- Applying the quadratic formula systematically to real-world scenarios
- Interpreting multiple solutions in context (ball going up vs. coming down)
- Converting between exact and approximate decimal answers
Visualizing the Motion
The ball follows a parabolic path, reaching 48 feet twice: once on the way up and once on the way down.
Solution: Method 1 — Direct Quadratic Formula Application
The height formula given appears to have a typo. The correct projectile motion formula is h = -16t² + v₀t (note the negative sign on the 16t² term, which represents gravitational acceleration).
Step 1 — Set up the equation
We know that v₀ = 96 ft/sec and we want to find when h = 48 feet. Substitute into the formula:
Step 2 — Rearrange to standard form
Move all terms to one side to get the equation equal to zero:
Step 3 — Simplify the coefficients
Divide the entire equation by -16 to make the numbers more manageable:
Step 4 — Apply the quadratic formula
For the equation at² + bt + c = 0, we have a = 1, b = -6, and c = 3. Using t = [-b ± √(b² - 4ac)] / (2a):
t = [6 ± √(36 - 12)] / 2
t = [6 ± √24] / 2
t = [6 ± 2√6] / 2
t = 3 ± √6
Step 5 — Calculate both solutions
Since √6 ≈ 2.449:
t₂ = 3 + √6 ≈ 3 + 2.449 = 5.449 seconds
Step 6 — Round to the nearest tenth
t₂ ≈ 5.4 seconds
Solution: Method 2 — Factoring by Completing the Square
Starting from our simplified equation t² - 6t + 3 = 0, we can solve by completing the square.
Step 1 — Move the constant term
Step 2 — Complete the square
Take half of the coefficient of t, square it, and add to both sides. Half of -6 is -3, and (-3)² = 9:
(t - 3)² = 6
Step 3 — Take the square root of both sides
t = 3 ± √6
This gives us the same result: t ≈ 0.6 seconds and t ≈ 5.4 seconds.
Verification
Let's check our answers by substituting back into the original height formula h = -16t² + 96t:
For t ≈ 0.551 seconds:
h = -16(0.304) + 52.896
h = -4.864 + 52.896 = 48.032 ≈ 48 feet ✓
For t ≈ 5.449 seconds:
h = -16(29.691) + 523.104
h = -475.056 + 523.104 = 48.048 ≈ 48 feet ✓
Common Pitfalls
Some students use
h = 16t² + v₀t instead of h = -16t² + v₀t. The coefficient of t² must be negative because gravity pulls the ball downward, creating a downward-opening parabola.
Students sometimes solve the quadratic but only report the smaller time value. In projectile motion, there are typically two times when an object reaches a given height: once on the way up and once on the way down.
The problem asks for answers to the nearest tenth. Some students give exact values like
3 - √6 or excessive decimal places like 0.5505102573 instead of rounding to 0.6 seconds.
When calculating
√(36 - 12), students sometimes compute √36 - √12 instead of √24. Remember that √(a - b) ≠ √a - √b.
The Physical Meaning
This problem illustrates the symmetric nature of projectile motion. The ball reaches 48 feet at two different times because:
- t ≈ 0.6 seconds: The ball is moving upward, gaining height
- t ≈ 5.4 seconds: The ball is moving downward, losing height
The total flight time (when the ball returns to ground level) would be when h = 0, giving us t = 0 (initial throw) and t = 6 seconds (landing). Notice that our two solutions at 48 feet are symmetric around the midpoint of 3 seconds, which is when the ball reaches its maximum height.
The General Pattern
For any projectile motion problem with the formula h = -16t² + v₀t, finding when height equals some value h₀ leads to:
16t² - v₀t + h₀ = 0
t = [v₀ ± √(v₀² - 64h₀)] / 32
The discriminant v₀² - 64h₀ tells us:
- If positive: two real times (ball reaches that height twice)
- If zero: one time (ball just touches that height at the peak)
- If negative: no real times (ball never reaches that height)
What If?
For a parabola h = -16t² + 96t, the vertex occurs at t = -b/(2a) = -96/(2(-16)) = 96/32 = 3 seconds.
Substitute t = 3: h = -16(3)² + 96(3) = -16(9) + 288 = -144 + 288 = 144 feet.
This makes sense: the maximum occurs at the midpoint between our two solutions (0.6 and 5.4 seconds), and 144 feet is well above our target height of 48 feet.
Answer: Maximum height is 144 feet at t = 3 seconds.
We know h = 32 at t = 1: 32 = -16(1)² + v₀(1)
32 = -16 + v₀, so v₀ = 32 + 16 = 48 ft/s.
Check: h = -16(1)² + 48(1) = -16 + 48 = 32 feet ✓
Answer: The initial velocity was 48 ft/s.
Set h = 0: 0 = -16t² + 96t
0 = t(-16t + 96) = t(-16)(t - 6) = -16t(t - 6)
t = 0 (initial throw) or t = 6 (landing)
At t = 6: h = -16(36) + 96(6) = -576 + 576 = 0 ✓
Answer: The ball hits the ground after 6 seconds.
60 = -16t² + 80t, so 16t² - 80t + 60 = 0
4t² - 20t + 15 = 0
t = [20 ± √(400 - 240)] / 8 = [20 ± √160] / 8 = [20 ± 4√10] / 8 = (5 ± √10) / 2
t₁ = (5 - √10)/2 ≈ 0.92 seconds, t₂ = (5 + √10)/2 ≈ 4.08 seconds
Difference = 4.08 - 0.92 = 3.16 ≈ 3.2 seconds
Answer: The time difference is 3.2 seconds.
Frequently Asked Questions
2026-06-28