Topic 7.6.1

1. Motion Tasks Topic 7.6.1

2. Topic 7.6.1 Motion Tasks California Standards: 23.0: Students apply quadratic equations to physical problems, such as the motion of an object under the force of gravity. What it means for you: You’ll model objects under the force of gravity using quadratic equations, and then solve the equations. • Key words: • quadratic • gravity • vertex • parabola • intercept • completing the square

3. Topic 7.6.1 Motion Tasks Quadratic equations have applications in real life. In particular, you can use them to model objects that are dropped or thrown up into the air.

4. Example 1 Topic 7.1.1 Motion Tasks The height of a stone thrown up in the air is modeled by the equation h = 80t – 16t2, where t represents the time in seconds since the stone was thrown and his the height of the stone in feet. After how many seconds is the stone at a height of 96 feet? Explain your answer. Solution The stone reaching a height of 96 feet is represented by h = 96, so you need to solve 80t – 16t2 = 96. Rewriting this in the form ax2 + bx + c = 0 (using t instead of x) gives: 16t2 – 80t + 96 = 0 t2 – 5t + 6 = 0 Divide through by 16 (t – 2)(t – 3) = 0 Factor the quadratic equation Solution follows… Solution continues…

5. Example 1 Topic 7.1.1 Motion Tasks The height of a stone thrown up in the air is modeled by the equation h = 80t – 16t2, where t represents the time in seconds since the stone was thrown and his the height of the stone in feet. After how many seconds is the stone at a height of 96 feet? Explain your answer. Solution (continued) (t – 2)(t – 3) = 0 t– 2 = 0 or t – 3 = 0 Solve using the zero property t = 2 or t = 3 So the stone is at a height of 96 feet after 2 seconds(on the way up), and again after 3 seconds(on the way down).

6. Topic 7.6.1 Motion Tasks Example 2 The height of a stone thrown up in the air is modeled by the equation h = 80t – 16t2, where t represents the time in seconds since the stone was thrown and his the height of the stone in feet. After how many seconds does the stone hit the ground? Explain your answer. Solution When the stone hits the ground, h = 0. So solve 80t – 16t2 = 0. 5t – t2 = 0 Divide through by 16 t(5 – t) = 0 t = 0 or t = 5 Solve using the zero property But t = 0 represents when the stone was thrown, so the stone must land at t = 5 — after 5 seconds. Solution follows…

7. Topic 7.6.1 Motion Tasks Example 3 The height of a stone thrown up in the air is modeled by the equation h = 80t – 16t2, where t represents the time in seconds since the stone was thrown and his the height of the stone in feet. Calculate h at t = 7. Explain your answer. Solution At t = 7, h = (80 × 7) – (16 × 72) = 560 – 784 = –224 Negative values of h suggest that the stone is beneath ground level. This can’t be true — the height can’t be less than zero feet. But the stone landed after 5 seconds — so after t = 5, the function h = 80t – 16t2doesn’t describe the motion of the stone. Solution follows…

8. 5 2 2 25 – = –16 t – 4 Topic 7.6.1 Motion Tasks Example 4 The height of a stone thrown up in the air is modeled by the equation h = 80t – 16t2, where t represents the time in seconds since the stone was thrown and his the height of the stone in feet.What is the maximum height of the stone? Justify your answer. Solution The graph of h = 80t – 16t2 is a parabola, and the maximum height reached is represented by the vertex of the parabola, which you can find by completing the square. h = 80t – 16t2 = –16(t2 – 5t) Solution follows… Solution continues…

9. 5 5 5 2 2 2 2 So the maximum height is 100 feet (which is reached at t = = 2.5 s). 25 – = –16 t – 4 2 + 100 = –16 t – Topic 7.6.1 Motion Tasks Example 4 The height of a stone thrown up in the air is modeled by the equation h = 80t – 16t2, where t represents the time in seconds since the stone was thrown and his the height of the stone in feet.What is the maximum height of the stone? Justify your answer. Solution (continued)

10. –5(t – 1)2 + 29 = 0 Þ (t – 1)2 = 29 5 Topic 7.6.1 Motion Tasks Guided Practice 1. In a Physics experiment, a ball is thrown into the air from an initial height of 24 meters. Its height h (in meters) at any time t (in seconds) is given by h = –5t2 + 10t + 24. Find the maximum height of the ball and the time t at which it will hit the ground. The ball’s maximum height is given by the vertex of the graph of h = –5t2 + 10t + 24 So complete the square: h = –5t2 + 10t + 24 = –5[t2 – 2t – 4.8] = –5[(t – 1)2 – 5.8] = –5(t – 1)2 + 29 So the vertex of the parabola is at (1, 29) Þ maximum height reached is 29 m. Ball lands when h = 0. So: Þ t – 1 = ±2.41 Þ t = 3.41 or t = –1.41 Maximum height is 29 meters and the ball hits the ground after 3.41 seconds of flight. Solution follows…

11. Topic 7.6.1 Motion Tasks Guided Practice 2. A firework is propelled into the air from the ground. Its height after tseconds is modeled by h = 96t – 16t2. The firework needs to explode at a height of 128 feet from the ground. After how long will it first reach this height? If the firework fails to explode, when will it hit the ground? Þ (t – 4)(t – 2) = 0 h = 128 ft Þ 96t – 16t2 = 128 Þ 16t2 – 96t + 128 = 0 So the height will be 128 feet at t = 2 (on the way up) and t = 4 (on the way down again) If it fails to explode, it will be on the ground again when 96t – 16t2 = 0 Þt2 – 6t = 0 Þt(t – 6) = 0 Þ t = 0 (this is when it was propelled) or t = 6 (this is when it lands) The firework will be at a height of 128 feet first after 2 seconds. If it fails to explode, it will hit the ground after 6 seconds. Solution follows…

12. 48 Topic 7.6.1 Motion Tasks Example 5 The height above the ground in feet (h) of a ball after t seconds is given by the quadratic function h = –16t2 + 32t + 48. Explain what the h-intercept and t-intercepts mean in this situation, and find the maximum height reached by the ball. Solution To get a clearer picture of what everything means, it helps to draw a graph. The intercept on the vertical axis(the h-axis) is found by putting t = 0: h = 48. Solution follows…

13. 48 3 –1 Topic 7.6.1 Motion Tasks Example 5 The height above the ground in feet (h) of a ball after t seconds is given by the quadratic function h = –16t2 + 32t + 48. Explain what the h-intercept and t-intercepts mean in this situation, and find the maximum height reached by the ball. Solution (continued) The intercepts on the horizontal axis (the t-axis) are found by solving h = 0 — that is, –16t2 + 32t + 48 = 0. t2 – 2t – 3 = 0 (t – 3)(t + 1) = 0 t = 3 or t = –1

14. 48 Topic 7.6.1 Motion Tasks Example 5 The height above the ground in feet (h) of a ball after t seconds is given by the quadratic function h = –16t2 + 32t + 48. Explain what the h-intercept and t-intercepts mean in this situation, and find the maximum height reached by the ball. Solution (continued) In this situation, the intercept on the vertical axis (the h-intercept) represents the initial height of the ball when it was thrown (at t = 0). So here, the ball was thrown from 48 feet above the ground. 3 –1

15. 48 Topic 7.6.1 Motion Tasks Example 5 The height above the ground in feet (h) of a ball after t seconds is given by the quadratic function h = –16t2 + 32t + 48. Explain what the h-intercept and t-intercepts mean in this situation, and find the maximum height reached by the ball. Solution (continued) The intercepts on the horizontal axis (the t-intercepts) represent the times at which the ball was at ground level. However, the function only describes the motion of the ball between t = 0 (when it was thrown) and t= 3 (when it lands). 3 –1

16. 48 Topic 7.6.1 Motion Tasks Example 5 The height above the ground in feet (h) of a ball after t seconds is given by the quadratic function h = –16t2 + 32t + 48. Explain what the h-intercept and t-intercepts mean in this situation, and find the maximum height reached by the ball. Solution (continued) So the t-intercept at t = 3 represents the point when the ball lands. The t-intercept at t = –1 doesn’t have any real-life significance here. To find the maximum height, you need to find the vertex of the parabola — so complete the square: 3 –1

17. (1, 64) 48 Topic 7.6.1 Motion Tasks Example 5 The height above the ground in feet (h) of a ball after t seconds is given by the quadratic function h = –16t2 + 32t + 48. Explain what the h-intercept and t-intercepts mean in this situation, and find the maximum height reached by the ball. Solution (continued) –16t2 + 32t + 48 = –16[t2 – 2t– 3] = –16[(t – 1)2 – 4] = –16(t – 1)2 + 64 The vertex of this parabola occurs where t = 1, and so the vertex is at (1, 64). This means the maximum height of the ball is 64 feet. 3 –1

18. Topic 7.6.1 Motion Tasks Independent Practice A baseball is hit from homebase. Its height in meters is modeled by the equation h = 25t – 5t2, where t is the time in seconds. 1. After how many seconds will the ball be at a height of 20 meters? 2. What height will the baseball reach before it starts descending? 1 second and 4 seconds 31.25 meters Solution follows…

19. Topic 7.6.1 Motion Tasks Independent Practice A rocket is fired into the air. Its height in feet at any time is given by the equation h = 1600t – 16t2, where t is the time in seconds. 3. Find the height of the rocket after 2 seconds. 4. After how many seconds will the rocket be 30,000 feet above the ground? 5. After how many seconds will the rocket hit the ground? 3136 feet After 25 seconds (on the way up), and after 75 seconds (on the way down). 100 seconds Solution follows…

20. seconds and 5 seconds 5 4 Topic 7.6.1 Motion Tasks Independent Practice 6. As a skydiver steps out of a plane, she drops her watch. The distance in feet, h, that the watch has fallen after t seconds is given by the equation h = 16t2 + 4t. After how many seconds will the watch have fallen 600 feet? The watch will have fallen 600 feet after 6 seconds. The height in feet of an object projected upwards is modeled by the equation h = 100t – 16t2. 7. How long after being projected is the object 100 feet above the ground? 8. What is the greatest height reached by the object? 156.25 feet Solution follows…

21. Topic 7.6.1 Motion Tasks Independent Practice The area of a rectangle is given by the formula A = x(20 – x) cm, where x is the width. 9. Find x when A = 84 cm2. 10. What value of x maximizes the area A? x = 14 cm or x = 6 cm x = 10 cm Solution follows…

22. Topic 7.6.1 Motion Tasks Independent Practice James and Mei are each standing on diving boards, and each throw a ball directly upwards. The height of each ball above the pool in feet, h, is plotted against the time in seconds, t, since it was thrown. 11. The height of James’s ball can be calculated using the equation h = –16t² + 30t + 10. From what height above the pool does James throw his ball? 10 feet 12. The height of Mei’s ball can be calculated using the equation h = –16t² + 32t + 20. After how many seconds does her ball reach its maximum height? 1 second 13. Calculate the difference in maximum heights of the balls, to 1 decimal place. Difference in heights = 11.9 feet Solution follows…

23. Topic 7.6.1 Motion Tasks Round Up Plotting a graph is often a good idea when you’re working on motion problems — because then you can see at a glance when an object reaches the ground, or when it reaches its maximum height.