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?
to find max height, plug your t = 2 into the original equation
s(2) = -16(4) + 64(2) + 200
= .....
at ground, s(t) = 0
-16t^2 + 64t + 200 = 0
4t^2 - 16t - 50 = 0
2t^2 - 8t - 25 = 0
use the quadratic formula:
t = (8 ± √264)/4
= appr 6.062 or a negative, which we will reject
verification:
http://www.wolframalpha.com/input/?i=s%28t%29%3D-16t%5E2%2B64t%2B200+
A) To find the maximum height of the ball, we need to determine the vertex of the quadratic function s(t) = -16t^2 + 64t + 200. The vertex of a quadratic function is given by the formula T = -b/2a, where T represents the time at which the maximum height occurs. In this case, a = -16 and b = 64, so we can substitute these values into the formula:
T = -(64) / 2(-16)
T = 64 / 32
T = 2
Therefore, the ball reaches its maximum height at 2 seconds.
To find the maximum height itself, we substitute the value of T back into the equation for s(t):
s(2) = -16(2)^2 + 64(2) + 200
s(2) = -64 + 128 + 200
s(2) = 264
So, the maximum height of the ball is 264 feet.
B) To determine when the ball hits the ground, we need to find the value of t when the height of the ball, s(t), is equal to 0. In other words:
-16t^2 + 64t + 200 = 0
To solve this quadratic equation, you can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -16, b = 64, and c = 200. Substituting these values into the quadratic formula, we have:
t = (-64 ± √(64^2 - 4(-16)(200))) / (2(-16))
t = (-64 ± √(4096 + 12800)) / (-32)
t = (-64 ± √(16896)) / (-32)
t ≈ (-64 ± 130) / (-32)
Now, solve for t using both the plus and minus signs:
t1 = (-64 + 130) / (-32)
t2 = (-64 - 130) / (-32)
Calculating these values, we get:
t1 ≈ 2.4
t2 ≈ -5
Since time cannot be negative in this context, we can disregard t2. Therefore, the ball hits the ground at around 2.4 seconds (rounded to the nearest tenth of a second).
C) To find s(0), we substitute t = 0 into the function s(t):
s(0) = -16(0)^2 + 64(0) + 200
s(0) = 200
Therefore, s(0) is equal to 200. This represents the initial height of the ball when it was first thrown. At t = 0, the ball is at a height of 200 feet above the ground.
D) To graph the quadratic function s(t) = -16t^2 + 64t + 200, you would plot points on a coordinate plane. Choose various values for t, substitute them into the function to find the corresponding values of s(t), and then plot those points. You can use a graphing calculator or software to automate this process.
Alternatively, you can find the vertex, maximum height, and x-intercepts (where the ball hits the ground) as mentioned earlier. These points will help you in sketching the graph accurately. The vertex represents the highest point on the graph, and the x-intercepts represent where the graph intersects the x-axis (i.e., when the ball hits the ground).