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

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.