9-4 Quadratic Equations and Projectiles A model rocket is shot at an angle into the air from the launch pad. 1. The height of the rocket when it has traveled horizontally x feet from the launch pad is given by h = - .163x² + 11.43x. A. Graph this equation x= -b 2a y x 15 25 35 45 55 70 y 135 184 200 136 1  200 150 100 50   x= -(11.43) 2(-.163)   x= -11.43 -.326 x 35   x 10 20 30 40 50 60 70 This is the x coordinate of the vertex

When the rocket is 10ft from the pad B. A 75 ft. tree, 10 ft. from the launch pad, is in the path of the rocket. Will the rocket clear the top of the tree? h = -.163(10)² + 11.43(10) h = -16.3 + 114.3 h = 98 When the rocket is 10ft from the pad it will be 98 ft. above the ground which will clear the 75 ft. tree. Sec. 9-4

2. The rockets height h at t seconds after launch is given by h = -22.2t² + 133t . -b 2a -133 2(-22.2) x =  3 = a. Graph this equation. b. How high is the rocket in 2 sec.? x 1 2 3 4 5 6 y  111  177  200 110  -1 y From the graph its  175 ft. Using the equation its: 200 150 100 50      h= -22.2t² + 133t -22.2(2)² + 133•2 -88.8 + 266 177.2 ft.   x 1 2 3 4 5 6 Sec. 9-4

General Formula for the Height of a Projectile over Time Let h be the height (in feet) of a projectile launched with an initial velocity v feet per second and an initial height of s feet. Then, after t seconds, h = -16t² + vt + s . Since 16ft ≈ 4.9 meters, if the units are in meters in the formula above, then h = -4.9t² + vt + s .

h = -16t² + vt + s h = -16t² + 64t + 10 b. h = -16(3)² + 64(3) + 10 1. A ball is thrown from an initial height of 10 feet with an initial velocity of 64 feet per second. a. Write an equation describing the height h in feet of the ball after t seconds. b. How high will the ball be after 3 seconds? c. What is the maximum height of the ball? h = -16t² + vt + s h = -16t² + 64t + 10 b. h = -16(3)² + 64(3) + 10 h = -144 + 192 + 10 h = 58 ft t = - b = -(64) = 2 2a 2(-16) h = -16(2)² + 64(2) + 10 h = -64 + 128 + 10 h = 74 ft

h = -4.9t² + vt + s h = -4.9t² + 40 b. h = -4.9t² + 40 0 = -4.9t² + 40 2. An object is dropped from an initial height of 40meters. a. Write a formula describing the height of the object (in meters) after t seconds. b. After how many seconds does the object hit the ground? c. What is the maximum height of the object? h = -4.9t² + vt + s h = -4.9t² + 40 b. h = -4.9t² + 40 0 = -4.9t² + 40 -40 - 40 -40/-4.9 = t² √-40/-4.9 = √t² √8.2 ≈ √t² 2.9sec. ≈ t t = - b = 0 = 0 2a 2(-4.9) h = -4.9(0)² + 40 h = 40m

t = - b = -30 ≈ 3 h = -4.9t² + vt + s 2a 2(-4.9) h = -4.9t² + 30t + 10 3. Suppose a ball is thrown upward with an initial upward velocity of 30 meters per second from an initial height of 10 meters. a Write a formula for the height in meters of the object after t seconds. b. Estimate when the ball is 40 meters high. t = - b = -30 ≈ 3 2a 2(-4.9) h = -4.9t² + vt + s h = -4.9t² + 30t + 10 b. 40 = -4.9t² + 30t + 10 80 ◙ ◙ window 0 < x < 10 0 < y < 80 1.2 4.9 1.2 sec and 4.9 sec 0 10