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Difficulty: MediumQuadratic Equations and the Quadratic Formula

A toy rocket is launched upward from a platform. Its height hh, in meters, above the ground tt seconds after launch is modeled by the function h(t)=−4.9t2+7.35t+12.25h(t) = -4.9t^2 + 7.35t + 12.25. According to this model, how many seconds after launch does the rocket strike the ground?

Answer: 2.5 seconds

Answer

The rocket strikes the ground 2.52.5 seconds after launch.
The correct answer of 2.52.5 is determined by setting the height h(t)h(t) to 00 and solving the resulting quadratic equation using the quadratic formula. Since time must be non-negative in this physical context, the negative solution of −1-1 is discarded, leaving 2.52.5 seconds as the time when the rocket strikes the ground.

Step-by-Step Solution

1
Set the height function h(t)h(t) to 00.
−4.9t2+7.35t+12.25=0-4.9t^2 + 7.35t + 12.25 = 0
The rocket strikes the ground when its height above the ground is 00 meters.
2
Apply the quadratic formula t=−b±b2−4ac2at = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
t=−7.35±(7.35)2−4(−4.9)(12.25)2(−4.9)t = \frac{-7.35 \pm \sqrt{(7.35)^2 - 4(-4.9)(12.25)}}{2(-4.9)}
This formula provides the solutions to any quadratic equation of the form at2+bt+c=0at^2 + bt + c = 0.
3
Calculate the discriminant and its square root.
b2−4ac=294.1225b^2 - 4ac = 294.1225 and 294.1225=17.15\sqrt{294.1225} = 17.15
Evaluating the term under the radical simplifies the quadratic formula expression.
4
Evaluate the two possible values for tt.
t=−1t = -1 or t=2.5t = 2.5
Solving the simplified expression gives the two mathematical roots of the quadratic equation.
5
Choose the physically valid solution.
t=2.5t = 2.5
Time must be positive in this scenario, so the negative solution t=−1t = -1 is discarded.

Key Concept

Solving a quadratic equation with decimal coefficients using the quadratic formula in a real-world motion context.
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Topics

Quadratic Equations and the Quadratic Formula

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