Algebra Quadratics Applications HW#54

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1 Algebra Quadratics Applications HW#54 1: A science class designed a ball launcher and tested it by shooting a tennis ball up and off the top of a 15-story building. They determined that the motion of the ball could be described by the function: h(t) = -16t t + 160, where t represents the time the ball is in the air in seconds and h(t) represents the height, in feet, of the ball above the ground at time t. a) Graph the function h(t) = -16t t (see below) b) What is the height of the building? c) At what time did the ball hit the The height of the building is also the height ground? of the tennis ball before it is launched into The ball hits the ground when the height is 0. the air. This occurs when t=0 so substitute 0 Therefore, we are looking for a solution to: for t and you get: -16t t = 0 H(0) = -16(0) (0) Use the quadratic formula or put this into a The height of the building is 160 feet. calculator. The solution is t=10 and -1, but only 10 makes sense. Therefore, the ball hits d) At what time did the ball reach its maximum height? You can put this into the calculator or you can realize that the maximum height is also the vertex. The x-value ( t in this case) is b 2a 144 which is = 4.5. (2)( 16) Therefore, the ball reached its maximum height at 4.5 seconds. the ground at 10 seconds. e) What is the maximum height of the ball? We calculated the time of the maximum height (4.5 seconds). Therefore, substitute 4.5 into the function to find the maximum height. -16(4.5) (4.5) The maximum height is 484 feet.

2 2: A rocket is launched from a cliff. The relationship between the height (h) in feet of the rocket and the time (t) in seconds since its launch can be represented by the following function: h(t) = -16t t a) When will the rocket hit the ground? The rocket hits the ground when the height is zero. -16t t = 0 Using the quadratic formula or using a calculator, we get t = -3 or 8. T = -3 makes no sense so the rocket hits the ground at 8 seconds. b) When will the rocket reach its maximum height? The maximum height is also the vertex. We use b or 80 which equals 2.5. Therefore, 2a 2( 16) the rocket reaches its maximum height at 2.5 seconds. c) What is the maximum height the rocket reaches? To get the maximum height, substitute 2.5 from part b into the function. We get: -16(2.5) (2.5) = 484 feet. d) At what height was the rocket launched? At the moment the rocket was launched, time (t) = 0. Therefore, substitute 0 for t and we get: -16(0) (0) = 384 feet. e) The function h(t) = -16t t is graphed. At which interval is the graph decreasing? The rocket is decreasing to the right of the vertex. Since the vertex is at time 2.5 seconds, the interval where the rocket is decreasing is all points when the time is greater than 2.5 (vertex) and less than 8 (when it hits the ground). The rocket is not decreasing at 2.5 or 8 seconds so we write the interval in one of two ways: 2.5 < t < 8 or (2.5, 8)

3 3: You have designed a new style of sports bike. Now you want to make a lot of them and sell them for profit. The profit is calculated by taking the amount that you sell the bike for and subtracting how much it cost you to make the bike. Profit = sale price cost to make The profit that you make in a month can be represented by the function: p = -20x x , where p represents the profit and x represents the sale price of one bicycle. a) At what price should you sell the bike to maximize your profits? The maximum is the vertex so use b 3400 or which equals 85. Therefore, in order to 2a 2( 20) maximize profits, we need to sell the bike for $85. b) Is there a price that will cause you to break even (meaning your profit will be exactly zero)? Set the function equal to zero and solve: -20x x = 0. You may simplify this equation by dividing both sides by -20: X 2 170x You may factor this or use the quadratic equation (or complete the square). Either way, you will get the solution: x = 150 or x = 20. Since both answers make sense, we include both of them. Therefore, we break even (profit = 0) when we sell the bike for $20 or $150.

4 4: A ball is shot from a cannon diagonally upward. The function that gives the height (in feet) of the ball based on the time is: h(t) = -16t t a) How high is the ball when it is sitting in the cannon (before it is shot)? When the ball is sitting in the cannon, time = 0. Therefore, substitute 0 for t and we get the height = 1.5 feet. b) Find the maximum height of the ball and at what time did it reach this height? The maximum height and the time when it reaches the maximum is the vertex of the parabola. Use b or 40 = 1.25 seconds. To get the maximum height, substitute a 2( 16) into the function and get 26.5 feet as the maximum height. c) How long does it take for the ball to reach the ground? (round to the nearest tenth of a second) The ball reaches the ground when the height is zero. Therefore, use the equation: -16t t = 0. Since it is not easy to factor this, we use the quadratic equation: 40 ± 1696 This simplifies to ± ( 16)(1.5) 2( 16) 1696 is not a perfect square so we normally would leave the answer like this. However, we are dealing with a real world problem so we must evaluate the square root. The problem says to round to the nearest tenth. Using a calculator, 1696 is rounded to Now we get: 40 ± or We can t have a negative time so we only consider the 32 solution , which is 2.5 when we round. Therefore, the ball will reach the ground at 2.5 seconds.

5 d) How long does it take the ball to reach a height of 20 feet? Rewrite the function, making the height 20 feet: -16t t = 20. Now solve by subtracting 20 from both sides and using the quadratic formula: -16t t 18.5 = 0 Using the quadratic formula we get: t =.6126 or Both of these make sense at first glance because the times are both positive. If the question were asking at what times is the ball 20 feet high?, the answer would be both. However, it is asking for how long it takes for the ball to reach 20 feet. This means we only want the first time. Therefore, the ball takes.6126 or.6 seconds (rounded) to reach 20 feet.