POLQ 3 | Lesson 4 | Solutions (Quadratic Word Problems)

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 \(x= \frac{-b}{2a}\)  which will give the x value for the line of symmetry and the max of minimum of any quadratic.

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

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

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

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:  \(x = \Large\frac{-b±\sqrt{b^2-4ac}}{2a}\)

a = -4, b = 40, c = 1

\(x = \Large\frac{-40 ±\sqrt{40^2 – 4(-4)(1)}}{2•(-4)}\)

\(x = \Large\frac{-40±\sqrt{1616}}{-8}\)

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

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.

 

 

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