If you are standing near the edge of the top of a 200 feet building and throw a ball vertically upward it will be modeled by the function s(t)=-16t^2+64t+200 where s(t) is the ball's height above ground in feet and t is seconds after the ball was thrown.
A) When does the ball reach its maximum height and what is the maximum height?
T= -b/2a
T= -64/-2(-16)
T= 2 seconds
How do I find the maximum height?
B) When does the ball hit the ground? (Round to the nearest tenth of a second)
Can you please help me with this question?
C) Find s(0) and describe what it represents?
Can you please help me with this question?
D) How would I graph the quadratic function
Please help me solving each parts step by step!!!!
I did this for you on Monday
http://www.jiskha.com/display.cgi?id=1449541115
A) To find the time when the ball reaches its maximum height, you can use the formula T = -b/2a, where a = -16 and b = 64 from the equation s(t) = -16t^2 + 64t + 200.
Plugging in the values, T = -64 / (2 * -16) = 2 seconds. So the ball reaches its maximum height at t = 2 seconds.
To find the maximum height, substitute t = 2 into the equation s(t) = -16t^2 + 64t + 200:
s(2) = -16(2)^2 + 64(2) + 200
= -64 + 128 + 200
= 264 feet
Therefore, the maximum height reached by the ball is 264 feet.
B) To find when the ball hits the ground, you need to determine the value of t when s(t) = 0. In other words, you need to solve the equation -16t^2 + 64t + 200 = 0.
You can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, it is easier to use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, you get:
t = (-64 ± √(64^2 - 4*(-16)*200)) / (2*(-16))
= (-64 ± √(4096 + 12800)) / (-32)
= (-64 ± √(16896)) / (-32)
≈ (-64 ± 130.005) / (-32)
Choosing the positive value (since time cannot be negative in this context), you have:
t ≈ (64 + 130.005) / 32
≈ 194.005 / 32
≈ 6.06 seconds (rounded to the nearest tenth)
Therefore, the ball hits the ground at approximately t = 6.1 seconds.
C) To find s(0), substitute t = 0 into the equation s(t) = -16t^2 + 64t + 200:
s(0) = -16(0)^2 + 64(0) + 200
= 0 + 0 + 200
= 200
So s(0) = 200. This represents the initial height of the ball when it is thrown. Therefore, s(0) = 200 feet.
D) To graph the quadratic function s(t) = -16t^2 + 64t + 200, you can follow these steps:
1. Determine the x-intercepts (where s(t) = 0) by solving the equation -16t^2 + 64t + 200 = 0 as mentioned in part B. In this case, we found t ≈ 6.1 seconds as the x-intercept.
2. Find the y-intercept by evaluating s(0). We found s(0) = 200. So the y-intercept is (0, 200).
3. Calculate the vertex, which represents the maximum height. We found the vertex occurs at t = 2 seconds, and s(2) = 264 feet. So the vertex is (2, 264).
4. Plot these key points: (0, 200), (2, 264), and (6.1, 0). You can also choose additional points to plot to help you draw the curve accurately.
5. Connect the plotted points smoothly to form a downward-opening parabolic curve. It represents the path of the ball's vertical motion.
Remember to label the axes (time on the x-axis and height on the y-axis) and add any necessary units to your graph.