So the dead line for my class is coming up. And I really need help for these questions!
2. Elaine shoots an arrow upward at a speed of 32 feet per second from a bridge that is 28 feet high. The height of the arrow is given by the function h(t) = -16t2+32t + 28, where t is the time in seconds.
a. What is the maximum height that the arrow reaches?
b. How long does it take the arrow to reach its maximum height?
c. How long would it take before the arrow reached the ground? Round your answer to the hundredths place.
3. A person standing close to the edge on the top of an 80-foot tower throws a ball with an initial speed of 64 feet per second. After t seconds, the height of the ball above the ground is
s(t) = -16t2 +64t + 80
a. After how many seconds will the ball reach its maximum height?
b. How long will it take before the ball reaches the ground?
c. What is the maximum height of the ball?
4. An object is launched at 19.6 meters per second from a 58.8 meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = -4.9t2+ 19.6t + 58.8, where s is in meters.
a. When does the object strike the ground?
b. What was its maximum height?
c. How long will it take to reach its maximum height?
5. A ball is thrown directly upward from an initial height of 200 feet with an initial velocity of 96 feet per second. After t seconds, the height of the ball above the ground is s(t) = 16t2+ 96t + 200.
a. After how many seconds will the ball reach its maximum height?
b. What was the maximum height?
c. How long will it take before the ball reaches the ground?
6. A soft-drink vendor at a popular beach analyzes his sales records, and finds that if he sells x cans of soda pop in one day, his profit (in dollars) is given by P(x) = -0.001x2+ 3x – 1800.
a. What is his maximum profit per day?
b. How many cans must be sold in order to obtain the maximum profit?
h(t)=-4.9t^2+24.5t+9.7
To answer these questions involving maximum height, maximum profit, and time taken, we will need to understand how to analyze quadratic functions and their graphs. The maximum or minimum value of a quadratic function occurs at its vertex.
1. The vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by the coordinates (-b/2a, f(-b/2a)).
a. To find the maximum height of the arrow, we need to find the vertex of the function h(t) = -16t^2 + 32t + 28. The time of maximum height can be found using the formula -b/2a. In this case, a = -16 and b = 32. Plugging these values into the formula, we get t = -32/(2(-16)) = 1.
Substituting t = 1 into the function h(t), we get h(1) = -16(1)^2 + 32(1) + 28 = 28 feet. Therefore, the maximum height the arrow reaches is 28 feet.
b. To find how long it takes for the arrow to reach its maximum height, we already found that the time is t = 1 second.
c. To find how long it takes for the arrow to reach the ground, we need to find the time when h(t) = 0. Setting -16t^2 + 32t + 28 = 0, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. The solutions are t ≈ 0.78 seconds and t ≈ 2.22 seconds. Rounding to the hundredths place, the arrow reaches the ground after approximately 0.78 seconds.
The same approach can be applied to questions 3, 4, and 5 to answer the parts a, b, and c.
2a. For the function s(t) = -16t^2 + 64t + 80, we need to find the vertex. Using the formula -b/2a, we get t = -64/(2(-16)) = 2 seconds. Substituting t = 2 into s(t), we get s(2) = -16(2)^2 + 64(2) + 80 = 144 feet. Therefore, the maximum height of the ball is 144 feet.
2b. To find how long it will take before the ball reaches the ground, we set s(t) = 0. Solving -16t^2 + 64t + 80 = 0, we find the solutions t ≈ 1.87 seconds and t ≈ 5.13 seconds. Rounding to the hundredths place, the ball reaches the ground after approximately 5.13 seconds.
6a. To find the maximum profit per day, we need to find the vertex of the function P(x) = -0.001x^2 + 3x - 1800. Using the formula -b/2a, we get x = -3/(2(-0.001)) = 1500. Substituting x = 1500 into P(x), we get P(1500) = -0.001(1500)^2 + 3(1500) - 1800 = $1200. Therefore, the maximum profit per day is $1200.
6b. To find the number of cans that must be sold to obtain the maximum profit, we look at the x-coordinate of the vertex, which is x = 1500. Therefore, he must sell 1500 cans of soda pop to obtain the maximum profit.
I hope this helps you solve these questions! Let me know if you need any further assistance.