1Cademy - Finding the Zeros and Specific Heights for $$h(t) = -16t^2 + 48t + 160$$

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How to Solve a Quadratic Equation by Factoring

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Finding the Zeros and Specific Heights for $h(t) = -16t^2 + 48t + 160$

To analyze the height of an object thrown upward, modeled by the polynomial function $h(t) = -16t^2 + 48t + 160$ , we can find its zeros, the time it reaches a specific height, and its height at a specific time. First, to find the zeros which indicate when the object hits the ground, set $h(t) = 0$ . Factoring the equation $-16t^2 + 48t + 160 = 0$ gives $-16(t^2 - 3t - 10) = 0$ , which factors further to $-16(t - 5)(t + 2) = 0$ . This yields $t = 5$ and $t = -2$ . Discarding the negative time, the object hits the ground at 5 seconds. Second, to find when the object is at a specific height of 160 feet, set $h(t) = 160$ . Subtracting 160 from both sides gives $-16t^2 + 48t = 0$ . Factoring gives $-16t(t - 3) = 0$ , yielding $t = 0$ and $t = 3$ . The object is at 160 feet at 0 and 3 seconds. Third, to find the height at a specific time, such as $t = 1.5$ seconds, substitute 1.5 into the function. The result is $h(1.5) = -16(1.5)^2 + 48(1.5) + 160 = 196$ . The object is at 196 feet after 1.5 seconds.

Updated 2026-05-25

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OpenStax Intermediate Algebra

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