# 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. 