) If you are standing near the edge of the top of a 200 ft. building and throw a ball
vertically upward it will be modeled by the function 𝑠(𝑡) = −16𝑡2 + 64𝑡 + 200 where
s(t) is the balls height above ground in feet and t is seconds after the ball was
thrown.
a. Graph the quadratic function
b. When does the ball reach its maximum height?
c. What is the maximum height?
d. When does the ball hit the ground? (round to the nearest tenth of a second)
e. Find s(0) and describe what it represents.
max height when t = -b/2a. That is, when t=2
Solve for t when h=0 to find when it hits the ground.
So, where is the ball when t=0? That is, where does it start?
a. To graph the quadratic function 𝑠(𝑡) = −16𝑡^2 + 64𝑡 + 200, we can plot a few points and connect them to form a smooth curve. Let's choose a few values of t and find the corresponding values of s(t).
For example, when t = 0, 𝑠(0) = −16(0)^2 + 64(0) + 200 = 200. So, one point on the graph is (0, 200).
Similarly, when t = 1, 𝑠(1) = −16(1)^2 + 64(1) + 200 = 248. Another point on the graph is (1, 248).
By finding a few more points in a similar manner, we can plot them on a graph and connect them to visualize the quadratic function.
b. The ball reaches its maximum height when its vertical velocity becomes zero. In other words, we need to find the time when the derivative of the function 𝑠(𝑡) equals zero. The derivative of 𝑠(𝑡) is given by 𝑠'(𝑡) = −32𝑡 + 64.
Setting 𝑠'(𝑡) = 0, we get −32𝑡 + 64 = 0. Solving this equation, we find that t = 2 seconds. Hence, the ball reaches its maximum height at t = 2 seconds.
c. To find the maximum height, we substitute the value of t = 2 back into the function 𝑠(𝑡). 𝑠(2) = −16(2)^2 + 64(2) + 200 = 216. Therefore, the maximum height reached by the ball is 216 feet.
d. The ball hits the ground when its height becomes zero. We can find this by setting s(t) = 0 in the function 𝑠(𝑡) and solving for t. -16𝑡^2 + 64𝑡 + 200 = 0.
Solving this quadratic equation, we can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / 2a. In this case, a = -16, b = 64, and c = 200.
Plugging in these values, we find that t ≈ 6.9 seconds. Hence, the ball hits the ground at approximately 6.9 seconds after it was thrown.
e. To find s(0), substitute t = 0 into the function 𝑠(𝑡). 𝑠(0) = −16(0)^2 + 64(0) + 200 = 200.
Therefore, s(0) represents the initial height of the ball above the ground when it was thrown, which in this case is 200 feet.