Chapter 1 - Equations and Inequalities - 1.5 Applications and Modeling with Quadratic Equations - 1.5 Exercises - Page 131: 36

Horizontal $30\ ft$, vertical $45\ ft$

Step 1. Based on the figure given in the exercise, use Pythagorean Theorem, we have: $x^2+(x+10)^2=(50)^2$ which gives $2x^2+20x+100=2500$ or $x^2+10x-1200=0$ Step 2. Factor $(x-30)(x+40)=0$ and solve the equation (discard the negative answer) to get $x=30\ ft$ Step 3. The vertical distance from the ground would be $x+10+5=45\ ft$