Solutions:

1. \(h(t) = 0.5 at^2+vt+s\) → \( h(t) = 0.5(-32)t^2+160(t)+4\) → \( h(t) = -16t^2+160t+4\)

2. \(h(3) = 0.5 at^2+vt +s\) → \(h(3) = 0.5(-32)3^2+160(3)+4\) → (h(3)=-144+480\) → \(h(3) = 360ft\)

3. Ball’s highest point can be found with \(x= \dfrac{-b}{2a}\) which will give the \(x\) value for the line of symmetry and the max of minimum of any quadratic.

\(h(t) = -16t^2+160t+4\)

\(t = -\dfrac{160}{2(-16)}\)

\(t = 5\); \(h(t) =-16\cdot5^2+160(5)+4\) →\(h(t) = 404\)

So, the highest point is (5, 404)

4. The answer is the larger answer as we want the \(x\)-coordinate of when the ball lands.

\(-16t^2+160t+4 = 0\)

Divide both sides by 4: \(-4t^2+40t+1= 0\)

To solve by quadratic formula: \(x = \dfrac{-b±\sqrt{b^2-4ac}}{2a}\)

\(a = -4, b = 40, c = 1\)

\(x = \dfrac{-40 \pm\sqrt{40^2-4(-4)(1)}}{2\cdot(-4)}\)

\(x = \dfrac{-40\pm\sqrt{1616}}{-8}\)

simplifying the radical \(x =\dfrac{-40±16\sqrt{101}}{-8}\)

simplifying and taking the largest value:

\(x = 5+2\sqrt{101}\) this is approximately 25 sec.

You could also find this by completing the square.

Graph below of intercept. Use a graphing utility such as Desmos to view whole graph.

Return to Practice (Quadratic Word Problems)

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