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1. 1. Quadratic Applications ------------------------------- Vertical Motion & Profit / Income By: Jeffrey Bivin Lake Zurich High School jeff.bivin@lz95.org Last Updated: November 30, 2007
2. 2. Vertical Motion • Compares the height of an object with the time in flight. 2 1 0 0 2 h(t ) = − gt + v t + h g = force of gravity: 32ft/sec or 9.8 m/sec vo = initial velocity ho = initial height Jeff Bivin -- LZHS
3. 3. A ball is thrown into the air from the top of a 200 foot tall building with an initial upward velocity of 80 ft/sec. • • • • • • • What is the maximum height of the ball? When will the ball reach the maximum height? When will the ball return to the ground? When will the ball be at a height of 250 feet? When will the ball be at a height of 400 feet? When will the ball be at a height of 50 feet? If the ball lands in a 20 foot deep pit, when will the ball hit the bottom of the pit? • What will be the height of the ball in 3 seconds? • How far from the building will the ball land? Jeff Bivin -- LZHS
4. 4. A ball is thrown into the air from the top of a 200 foot tall building with an initial upward velocity of 80 ft/sec. • What is the maximum height of the ball? We need to use: h(t ) = − 1 gt 2 + v0t + h0 2 g = 32 ft/s2 vo = 80 ft/s ho = 200 ft h(t ) = − (32)t + 80t + 200 h(t ) = − 16t 2 + 80t + 200 1 2 Jeff Bivin -- LZHS 2
5. 5. A ball is thrown into the air from the top of a 200 foot tall building with an initial upward velocity of 80 ft/sec. • What is the maximum height of the ball? h(t ) = − 16t 2 + 80t + 200 Where is the maximum? t t t t = = = = −b 2a −80 2 ( −16 ) −80 −32 5 2 Jeff Bivin -- LZHS h( Find the vertex…… ) = − 16( ) + 80( 5 ) + 200 2 h( 5 ) = − 16( 25 ) + 80( 5 ) + 200 2 4 2 h( 5 ) = 300 2 ( t , h(t ) ) Vertex is: ( 5 , 300 ) 2 5 2 5 2 2 300 ft.
6. 6. A ball is thrown into the air from the top of a 200 foot tall building with an initial upward velocity of 80 ft/sec. • When will the ball reach the maximum height? 2.5 sec h(t ) = − 16t 2 + 80t + 200 Where is the maximum? t t t t = = = = −b 2a −80 2 ( −16 ) −80 −32 5 2 Jeff Bivin -- LZHS h( Find the vertex…… ) = − 16( ) + 80( 5 ) + 200 2 5 5 2 h( 2 ) = − 16( 2 ) + 80( 5 ) + 200 2 h( 5 ) = 300 2 ( t , h(t ) ) Vertex is: ( 5 , 300 ) 2 5 2 5 2 2
7. 7. A ball is thrown into the air from the top of a 200 foot tall building with an initial upward velocity of 80 ft/sec. • When will the ball return to the ground? 6.830 sec. h(t ) = − 16t 2 + 80t + 200 What is the height at the ground? h(t) = 0 0 = − 16t + 80t + 200 2 t= t= −80 ± 80 2 − 4 ( −16 )( 200 ) 2 ( −16 ) −80 ± 19200 −32 Get the decimal approximations: Jeff Bivin -- LZHS t ≈ − 1.830 t ≈ 6.830
8. 8. A ball is thrown into the air from the top of a 200 foot tall building with an initial upward velocity of 80 ft/sec. • When will the ball be at a height of 250 feet? 2 h(t ) = − 16t + 80t + 200 0.732 sec 250 feet What height? h(t) = 250 & 2 4.268 sec. 250 = − 16t + 80t + 200 0 = − 16t + 80t − 50 2 t= t= −80 ± 80 2 − 4 ( −16 )( −50 ) 2 ( −16 ) −80 ± 3200 −32 Get the decimal approximations: Jeff Bivin -- LZHS t ≈ 0.732 t ≈ 4.268
9. 9. A ball is thrown into the air from the top of a 200 foot tall building with an initial upward velocity of 80 ft/sec. • When will the ball be at a height of 400 feet? never h(t ) = − 16t 2 + 80t + 200 What height? h(t) = 400 400 = − 16t 2 + 80t + 200 0 = − 16t 2 + 80t − 200 t= t= −80 ± 80 2 − 4 ( −16 )( − 200 ) 2 ( −16 ) −80 ± −6400 −32 Wait, what was the maximum height? Jeff Bivin -- LZHS 300 ft
10. 10. A ball is thrown into the air from the top of a 200 foot tall building with an initial upward velocity of 80 ft/sec. • When will the ball be at a height of 50 feet? 6.453 sec. h(t ) = − 16t 2 + 80t + 200 What height? h(t) = 50 50 = − 16t 2 + 80t + 200 0 = − 16t 2 + 80t + 150 t= t= −80 ± 80 2 − 4 ( −16 )( 150 ) 2 ( −16 ) −80 ± 16000 −32 Get the decimal approximations: Jeff Bivin -- LZHS t ≈ − 1.453 t ≈ 6.453
11. 11. A ball is thrown into the air from the top of a 200 foot tall building with an initial upward velocity of 80 ft/sec. • If the ball lands in a 20 foot deep pit, when will h(t ) the ball 2hit thet bottom of the pit? = − 16t + 80 + 200 What height? h(t) = -20 − 20 = − 16t 2 + 80t + 200 20 feet below the ground Jeff Bivin -- LZHS
12. 12. A ball is thrown into the air from the top of a 200 foot tall building with an initial upward velocity of 80 ft/sec. • If the ball lands in a 20 foot deep pit, when will the ball hit the bottom of the pit? 6.972 sec. 2 h(t ) = − 16t + 80t + 200 What height? h(t) = -20 − 20 = − 16t 2 + 80t + 200 0 = − 16t 2 + 80t + 220 t= t= −80 ± 80 2 − 4 ( −16 )( 220 ) 2 ( −16 ) −80 ± 20480 −32 Get the decimal approximations: Jeff Bivin -- LZHS t ≈ − 1.972 t ≈ 6.972
13. 13. A ball is thrown into the air from the top of a 200 foot tall building with an initial upward velocity of 80 ft/sec. • What will be the height of the ball in 3 seconds? 296 ft. h(t ) = − 16t 2 + 80t + 200 What time? t=3 h(3) = − 16(3) 2 + 80(3) + 200 h(3) = − 144 + 240 + 200 h(3) = 296 Jeff Bivin -- LZHS
14. 14. A ball is thrown into the air from the top of a 200 foot tall building with an initial upward velocity of 80 ft/sec. • How far from the building will the ball land? h(t ) = − 16t + 80t + 200 2 Wait !!!! This formula compares time with height, not horizontal distance. Answer: Jeff Bivin -- LZHS we don’t know!
15. 15. Jeff Bivin -- LZHS
16. 16. A diver dives off a 3 meter diving board into a pool with an initial upward velocity of 3.5 m/sec. • What is the maximum height of the diver? • When will the diver reach his/her maximum height? • When will the diver splash into the water? • What will be the height of the diver in 1 second? Jeff Bivin -- LZHS
17. 17. A diver dives off a 3 meter diving board into a pool with an initial upward velocity of 3.5 m/sec. • What is the maximum height of the diver? 3.625 meters • When will the diver reach his/her maximum height? 0.358 sec. • When will the diver splash into the water? 1.217 sec. • What will be the height of the diver in 1 second? 1.6 meters Jeff Bivin -- LZHS
18. 18. Jeff Bivin -- LZHS
19. 19. A taxi service operates between two airports transporting 200 passengers a day. The charge is \$15.00. The owner estimates that 10 passengers will be lost for each \$2 increase in the fare. What charge would be most profitable for the service? What is the maximum income? VERTEX Define the variable x = number of \$2 price increases f(x) = income x= −b 2a x= −250 2 ( −20 ) x = 6.25 Income = Price ● Quantity f(x) = ( 15 + 2x ) ( 200 – 10x ) f(x) = 3000 – 150x + 400x – 20x2 f(x) = – 20x2 + 250x + 3000 f(6.25) = – 20(6.25)2 + 250(6.25) + 3000 Vertex is: f(6.25) = 3781.25 ( 6.25, 3781.25) So, price = (15 + 2x) = (15 + 2(6.25)) = 15 + 12.5 = \$27.50 Maximum income = f(x) = \$3781.25 Jeff Bivin -- LZHS
20. 20. Jeff Bivin -- LZHS

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