A biker traveling with a velocity of 80 feet per second leaves a 100 foot platform and is projected directly upward. The function modeling the projectile motion of the biker is s(t)=-16t^2+80t+100,where s(t) is the height and t is the number of
seconds the biker is in the air.
a) Draw a rectangular coordinate system and sketch the height of the biker
after the bike leaves the platform. Make the horizontal axis the time the
biker is in the air. Label the horizontal and vertical axes. Don’t forget that
the biker leaves the platform at 100 feet.
b) Use your graphing utility to graph the parabola.
c) Using your graphing calculator, find how many seconds it takes for the biker
to reach its maximum height. Compare this value to –b/2a, where a and b
are the coefficients found from comparing the form s(t)=at^2+bt+c to the given quadratic function. This value -b/2a, is the t-value of the vertex. Find the y-value at this point by evaluating the function at the t-value of -b/2a.
What does the y-value represent at this specific t-value? Then state the (t,s(t)) point of the vertex or maximum point.
d) Looking at your calculator, find the number of seconds the biker is in the air
(or “hang time”). Then find the range of height values that the biker attains.
e) State the domain and range of this real life example. Remember that time
isn’t negative and that the model is valid only when the biker is in the air!
f) Use the graph to determine when the biker will reach a height of 100 feet.
State the (these) point(s) as an order pair and using functional notation.
g) Use the graph to determine what height the biker will attain after 1 second.
State the point as an ordered pair and using functional notation.
h) Where is the function increasing? In other words, for what t-values does the
biker continue to get higher? Where is the function decreasing? In other
words, for what t-values does the biker start descending toward the ground?
State these intervals using interval notation.
T = Time in seconds.
h = Ht. in feet.
(T, h)
(0,100) Platform.
(0.5,136)
(1.0,164)
(1.5,184)
(2.0,196)
V(2.5,200) Max. point.
(3.0,196)
(3.5,184)
(4.0,164)
(4.5,136)
(5.0,100). Platform.
a) To sketch the height of the biker after the bike leaves the platform, we will plot a graph with time on the horizontal axis and height on the vertical axis. Since the biker leaves the platform at a height of 100 feet, we will label the vertical axis as "Height (feet)" and the horizontal axis as "Time (seconds)". We will start the time axis from 0 and extend it based on the problem's context.
b) Use a graphing utility (such as an online graphing calculator or software) to graph the parabola given by the equation s(t) = -16t^2 + 80t + 100. Set the vertical axis to represent the height (s(t)) and the horizontal axis to represent the time (t). Plot points on the graph by substituting different values of t into the equation and calculating the corresponding s(t) values.
c) Using the graphing calculator, find the number of seconds it takes for the biker to reach their maximum height. This can be done by finding the t-value of the vertex of the parabola. This value can be compared to -b/2a, where a and b are the coefficients from the quadratic function. In this case, a = -16 and b = 80. So, -b/2a = -80 / (2 * (-16)) = 5 seconds.
To find the corresponding height at this time, evaluate the function s(t) at t = 5. So, s(5) = -16(5)^2 + 80(5) + 100 = 300 feet. The y-value at the vertex represents the maximum height reached by the biker. The vertex or maximum point is (5, 300).
d) Looking at the graph, find the number of seconds the biker is in the air, which is also known as the "hang time". The time at which the biker reaches the ground will be the duration of the entire flight. From the graph, read the horizontal intercept(s) or the x-values where the graph intersects the x-axis. These x-values represent the time(s) when the height is 0, indicating that the biker has landed. The range of height values that the biker attains can be determined by looking at the y-values on the graph.
e) The domain of this real-life example is the set of all non-negative real numbers, since time cannot be negative. The model is valid only when the biker is in the air. So, the domain is [0, ∞).
The range of this real-life example is the set of all real numbers greater than or equal to 100 (assuming the biker starts from a height of 100 feet). As the biker's height changes during the flight, the function can take on any value greater than or equal to 100. So, the range is [100, ∞).
f) To determine when the biker will reach a height of 100 feet, we need to find the point(s) on the graph where s(t) = 100. These points will represent the time(s) when the height is 100 feet. You can either look for the intersection of the graph with the horizontal line y = 100 or solve the equation -16t^2 + 80t + 100 = 100. The ordered pair or functional notation for this point will be (t, s(t)) = (t, 100).
g) To find the height the biker will attain after 1 second, we need to find the point on the graph where t = 1. We can either read this point directly from the graph or substitute t = 1 into the equation s(t) to find s(1) = -16(1)^2 + 80(1) + 100. The ordered pair or functional notation for this point will be (t, s(t)) = (1, s(1)) = (1, height value).
h) To determine where the function is increasing or decreasing, we need to find the intervals of t-values where the biker continues to get higher (increasing) and the intervals where the biker starts descending toward the ground (decreasing). From the graph, identify the sections where the graph is rising (increasing) and falling (decreasing). The intervals can be expressed using interval notation. For example, if the biker is increasing from t = 2 to t = 5, and decreasing from t = 5 to t = 8, the increasing interval is (2, 5) and the decreasing interval is (5, 8).