﻿ Real World applications of Parabolas. Several practice problems with pictures and answers

# Parabola in Real Life

Real world applications and problems

### Practice Problems

##### Problem 1

Vertex of parabola is (0,40)

The helicopter must be above 40.

This real is simply a real world application of how to find the vertex of a parabola

##### Problem 2

The ball lands at the solution of this quadratic equation. There are two solutions. One at 2 and the other at − 2. This picture assumes that Joseph threw the ball to the right so that the whiffle balls lands at 2

You can solve this quadratic by factoring or by using the quadratic formula

##### Problem 3

The ball lands at the solution of this quadratic equation. There are two solutions. One at 2 and the other at − 2. This picture assumes that Joseph threw the ball to the right so that the whiffle balls lands at 1.

##### Problem 4

The ball hits the ground at d = 0. To find the value of t at this point we must solve this quadratic equation.

0 = −t² + 36
t² = 36
t = 6

(Note: t = -6 is also a solution of this equation. However, only the positive solution is valid since we are measuring seconds.)
##### Problem 5

We want to find when d=0, which represents the moment when the ball hits the ground.

d = 0, when 0 = -5t² +60
5t² = 60
t² = 60 ÷5 = 12

$$\sqrt{12} \approx 3.5$$

##### Problem 6
Step 1

Identify all of the occurrences of 't' and substitute the input in

$$h( {\color{blue} t} ) = -5 {\color{blue} t} ^2 + 41 {\color{blue}t} + 1.2 \\ h( {\color{blue}4} ) = -5 \cdot {\color{blue}4}^2 + 40 \cdot {\color{blue} 4} + 1.2$$

Step 2

Compute result

$$h( {\color{blue} 4}) = -5 \cdot {\color{blue} 4}^2 + 40 \cdot {\color{blue} 4} + 1.2 \\ h( {\color{blue} 4}) = {\color{red} 81.2}$$

Step 3

$$h({\color{blue} input}) = {\color{red} output} \\ h({\color{blue} 4} ) = {\color{red} 81.2}$$

Here is a picture of graph of projectile's path with the point $$({\color{blue} t}, {\color{red} h(t)}) ({\color{blue} 4}, {\color{red} 81.2})$$