Question 2.6.9: A projectile is fired vertically upward from a height of 600......
A projectile is fired vertically upward from a height of 600 feet above the ground. Its height h(t) in feet above the ground after t seconds is given by
h(t)=-16t²+803t+600
(a) Determine a reasonable viewing rectangle that includes all pertinent features of the graph of h.
(b) Estimate when the height of the projectile is 5000 feet above the ground.
(c) Determine when the projectile will be more than 5000 feet above the ground.
(d) How long will the projectile be in flight?
Height of the projectile above the ground after t seconds: h(t) = -16t² + 803t + 600
Initial height of the projectile above the ground: 600 feet
Height of the projectile when it is 5000 feet above the ground: 5000 feet
Height at which the projectile is more than 5000 feet above the ground: more than 5000 feet
Unknown time duration of the projectile’s flight.
Final Answer
(a) The graph of h is a parabola that opens downward. To estimate Ymax (note that we use x and y interchangeably with t and h), let us approximate the maximum value of h. Using
t=-\frac{b}{2 a}=-\frac{803}{2(-16)} \approx 25.1,we see that the maximum height is approximately h(25)=10,675.
The projectile rises for approximately the first 25 seconds, and because its height at t=0,600 feet, is small in comparison to 10,675 , it will take only slightly more than an additional 25 seconds to fall to the ground. Since h and t are positive, a reasonable viewing rectangle is
Calculator Note: Once we determine the Xmin and Xmax values, we can use the ZoomFit feature to graph a function over the interval [Xmin, Xmax]. In this example, assign 0 to Xmin and 51 to Xmax and then select ZoomFit under the ZOOM menu.
(b) We wish to estimate where the graph of h intersects the horizontal line h(t)=5000, so we make the assignments
and obtain a display similar to Figure 10 . It is important to remember that the graph of Y_1 shows only the height at time t— it is not the path of the projectile, which is vertical. Using an intersect feature, we find that the smallest value of t for which h(t)=5000 is about 6.3 seconds.
Since the vertex is on the axis of the parabola, the other time at which h(t) is 5000 is approximately 25.1-6.3, or 18.8 , seconds after t=25.1—that is, at t \approx 25.1+18.8=43.9 \mathrm{sec}.
(c) The projectile is more than 5000 feet above the ground when the graph of the parabola in Figure 10 is above the horizontal line—that is, when
(d) The projectile will be in flight until h(t)=0. This corresponds to the x-intercept in Figure 10 . Using a root or zero feature, we obtain t \approx 50.9 \mathrm{sec}. (Note that since the y-intercept is not zero, it is incorrect to merely double the t value of the vertex to find the total time of the flight; however, this would be acceptable for problems in which h(0)=0.)
