Example

Finding the Time for a Firework to Reach 260 Feet

To find when a projectile reaches a given height, substitute the known values into the projectile motion formula (h=16t2+v0th = -16t^2 + v_0 t) and solve for time (tt).

Problem: A firework is shot upward with an initial velocity of 130 feet per second. How many seconds will it take to reach a height of 260 feet? Round to the nearest tenth of a second.

Solution: Substitute the initial velocity (v0=130v_0 = 130) and target height (h=260h = 260) into the formula: 260=16t2+130t260 = -16t^2 + 130t

Rewrite this quadratic equation in standard form by moving all terms to one side: 16t2130t+260=016t^2 - 130t + 260 = 0

Identify the coefficients (a=16a = 16, b=130b = -130, c=260c = 260) and substitute them into the Quadratic Formula: t=(130)±(130)24(16)(260)2(16)t = \frac{-(-130) \pm \sqrt{(-130)^2 - 4(16)(260)}}{2(16)}

Simplify the expression: t=130±16,90016,64032t = \frac{130 \pm \sqrt{16{,}900 - 16{,}640}}{32} t=130±26032t = \frac{130 \pm \sqrt{260}}{32}

Approximate the two solutions: t=130+260324.6ort=130260323.6t = \frac{130 + \sqrt{260}}{32} \approx 4.6 \quad \text{or} \quad t = \frac{130 - \sqrt{260}}{32} \approx 3.6

Unlike geometry applications where negative or secondary solutions might be discarded, both positive time values are valid here. The firework reaches 260 feet after approximately 3.6 seconds on the way up, and passes that height again at 4.6 seconds on the way down.

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Updated 2026-06-26

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