Lesson 17 Quadratic Word Problems. The equation to model Vertical Motion is

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W8D1 Quadratic Word Problems Warm Up 1. A rectangle has dimensions of x+2 and x+3. What is the area of the rectangle? 2. What is the Perimeter of the rectangle? 3. If the area of the rectangle is 30 m 2, what is x? Lesson 17 Quadratic Word Problems The equation to model Vertical Motion is h(t) = 5t 2 + v o t + h 0 h(t)is the height above the ground after a given time tthe time since launch 5the constant to represent gravity (if the problem units are meters) v 0 the initial velocity h 0 the initial height Once you set the equation for a problem don t change any of the constants! EX 1: A twirler throws a baton with an initial upward velocity of 20 m/s and initial height of 4 m a. Calculate its height after 2 sec. b. When will it be 19 m above the ground? c. When will the baton be back to the twirler s level? d. When will the baton be at the maximum height? What is that height? e. Sketch a graph of the situation plotting all of the points that you found in parts a, b,c and d. Equation for this problem: h(t) = 5t 2 + 20t + 4 h(2) = 24ft b. 19 = 5t 2 + 20t + 4 0 = 5t 2 + 20t 15 0 = 5(t 2 4t + 3) 0 = 5(t 3)(t 1) t = 1, 3sec c. 4 = 5t 2 + 20t + 4 0 = 5t 2 + 20t 0 = 5t(t 4) t = 4sec maximum occurs at the vertex. Solve for the a.o.s then vertexd. b 2a = 20 10 = 2sec 2 seconds, 24 ft

25 Vertex (2 sec, 24 ft) 23 21 19 17 15 13 11 9 7 5 3 y-intercept (0 sec, 4 ft) 1 1 1 2 3 4 5 6 7 8 9 10 Figure 1: Example1h(t) = 5t 2 + 20t + 4

EX 2: 7. At the circus, Art Tillery is fired into the air from a cannon on a platform. His initial upward velocity is 7 m/s a. How high is he above the firing point after 0.6 sec? b. When will be be 2 m above the platform? c. When will he be back to the level of the canon? d. When will he land in the tank of water 20 m below his firing point? h(t) = 5t 2 + 7t h(.6) = 5(.6) 2 + 7(.6) h(.6) = 2.4 b. 2 = 5t 2 + 7t 0 = 5t 2 + 7t 2 0 = (5t 2)(t 1) t = 2 5 1sec c. 0 = 5t 2 + 7t 0 = t(5t 7) t = 7 5 sec d. 20 = 5t 2 + 7t 5t 2 + 7t + 20 = 0 CAN T FACTOR t = 2.818 EX 3: You go to the roof of a twelve-story building and look over the edge to the reflecting pool 50 m below. Your friend drops their book over the edge at the same instant that you chuck your book straight down at 15 meters per second. How long until each book hits the ground? Your velocity is going down so it is NEGATIVE 15. Your friend doesn t throw the book at all so velocity is zero. Friend h(t) = 5t 2 + 0t + 50 You h(t) = 5t 2 15t + 50 You: -5(t 2 +3t-10) -5 (t+5)(t-2) t = 2 seconds Friend: -5t 2 +50 = 0-5t 2 = -50 t 2 = 10 t = 10 3.16 sec

1. a. h(t) = 5t 2 + 25t + 0 h(2) = 30ft h(3) = 30ft b. 20 = 5t 2 + 25t t = 1, 4sec c.graph d.0 = 5t(t 5) t = 5sec Exit Pass 1. A paintball sniper is up in the stadium 64 feet above the ground. It shoots out a paintball at 48 feet per second, at the enemy team. How high is this ball after 3 seconds? 2. How long will it take for the ball to hit the ground? h(t) = 16t 2 + v 0 t + h 0 h(t) = 16t 2 + 48t + 64 h(t) = 16(t 2 3t 4) h(t) = 16(t 4)(t + 1) t = 4sec

Extra Examples (units = feet) EX 1: A batter hits a pop fly. The function h(t) = 16t 2 + v 0 t + h 0 represents the height of the baseball, in feet, at t seconds. h(t) = 16t 2 + v 0 t + h 0 if the units are feet h(t) = 5t 2 + v 0 t + h 0 if the units are meters Coefficient of gravity : -16 or -5 depending on the units. Always multiplied by t 2 t = time h(t) = height at a certain time v 0 = starting velocity. Velocity changes over time! h 0 = starting height Define the variables then plug into the eq uation How high is the ball after 3 seconds if it is hit at a speed of 96 feet/sec from 4 feet off the ground? h(t) = 16t 2 + v 0 t + h 0 h(t) = 16t 2 + 96t + 3 h(3) = 16(3) 2 + 96(3) + 4 h(3) = 148 When does it hit the ground? h(t) = 16t 2 + v 0 t + h 0 0 = 16t 2 + 96t + 4 this doesn t factor so for now you would just stop here 3 = 16(x 2 6x+) 144 3 = 16(x 4) 2 147 = (x 4)2 16 147 x = 4 ± 4 x =.031or6.031 An object is released into the air at an initial height of 6 feet and an initial velocity of 30 feet per second. The object is caught at a height of 7 feet. Use the vertical motion model, where h is the height, t is the time in motion, s is the initial height, and v is the initial velocity, to find how long the object is in motion. Throwing an Object on the Moon An astronaut standing on the moon throws a rock upwards with an initial velocity of 27 feet per second. The astronaut?s hand is 6 feet above the surface of the moon. The height of the rock is given by h= 2.7t 2 + 27t + 6 How many seconds does it take for the rock to fall to the ground? Suppose the astronaut had been standing on Earth. Write a vertical motion model for the height of the rock after it is thrown. how many seconds does it takes for the rock to fall to the ground on Earth.

We want to construct a box whose base length is 3 times the base width. The material used to build the top and bottom cost 10/f t2andthematerialusedtobuildthesidescost6/ft2. If the box must have a volume of 50 ft3 determine the dimensions that will minimize the cost to build the box. What is the maximum area for a rectangle with perimeter of 200 ft P = 200ft Rectangle x by y with semicircle ends. a. r = y 2 68. b. d = 1 πy each 2 c. 2x + πy = 200 y = d. A rect = 200x 2x2 π 200 2x π Find the vertex using b 2a x = 50, y = 1590 x y = 1 3 (200 4x) 2xy 2 64 256 4 61.33 490.67 10 53.33 1066.67 20 40 1666.67 25 25 1066.67 A = 2 3 (200x 4x2 ) A = 8 3 x2 + 400 3 x vertex: 400 16 = 25, (25, 1666)