Example 2: A volleyball is served into the air at 30 ft/sec from a height of 4 ft. The quadratic equation h = -16t² + 30t + 4 represents the height in feet of the ball over time in seconds. The graph shows the parabolic trajectory of the ball.

We are given a quadratic equation that models the height of a volleyball over time:

> h = –16t² + 30t + 4

Where:
- h is the height in feet,
- t is the time in seconds after the serve.

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🎯 Goal:

Find how long it takes for the ball to hit the floor — that is, when h = 0.

So we need to solve the equation:

> –16t² + 30t + 4 = 0

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This is a quadratic equation in standard form:
at² + bt + c = 0, where
- a = –16
- b = 30
- c = 4

We’ll use the quadratic formula:

> t = [–b ± √(b² – 4ac)] / (2a)

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🔢 Step-by-step calculation:

1. Compute discriminant (D):
D = b² – 4ac
D = (30)² – 4(–16)(4)
D = 900 – (–256)
D = 900 + 256 = 1156

2. Take square root of discriminant:
√1156 = 34 *(since 34 × 34 = 1156)*

3. Plug into quadratic formula:

t = [–30 ± 34] / (2 × –16)
t = [–30 ± 34] / (–32)

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📌 Two possible solutions:

First solution (using +):
t = (–30 + 34) / (–32) = 4 / (–32) = –0.125

Second solution (using –):
t = (–30 – 34) / (–32) = (–64) / (–32) = 2

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✔ Interpretation:

Time cannot be negative in this context, so we discard t = –0.125.

The only meaningful solution is t = 2 seconds.

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🏁 Final Answer:

> It will take 2 seconds for the volleyball to hit the floor.

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💡 Extra Insight:

The graph shows a parabola opening downward (because a = –16 < 0), which makes sense — the ball goes up, reaches a peak, then falls down. The x-intercepts represent times when height = 0. One is before the serve (negative time, not physical), and the other is at t = 2 seconds — when it lands.

✔ So, 2 seconds is the correct and only physically meaningful answer.