Answer:
Part A) The graph in the attached figure (see the explanation)
Part B) 16 feet
Part C) see the explanation
Step-by-step explanation:
Part A) Graph the function
Let
h(t) ----> the height in feet of the ball above the ground
t -----> the time in seconds
we have
[tex]h(t)=-16t^{2}+98[/tex]
This is a vertical parabola open downward (the leading coefficient is negative)
The vertex is a maximum
To graph the parabola, find the vertex, the intercepts, and the axis of symmetry
Find the vertex
The function is written in vertex form
so
The vertex is the point (0,98)
Find the y-intercept
The y-intercept is the value of the function when the value of t is equal to zero
For t=0
[tex]h(t)=-16(0)^{2}+98[/tex]
[tex]h(0)=98[/tex]
The y-intercept is the point (0,98)
Find the t-intercepts
The t-intercepts are the values of t when the value of the function is equal to zero
For h(t)=0
[tex]-16t^{2}+98=0[/tex]
[tex]t^{2}=\frac{98}{16}[/tex]
square root both sides
[tex]t=\pm\frac{\sqrt{98}}{4}[/tex]
[tex]t=\pm7\frac{\sqrt{2}}{4}[/tex]
therefore
The t-intercepts are
[tex](-7\frac{\sqrt{2}}{4},0), (7\frac{\sqrt{2}}{4},0)[/tex]
[tex](-2.475,0), (2.475,0)[/tex]
Find the axis of symmetry
The equation of the axis of symmetry in a vertical parabola is equal to the x-coordinate of the vertex
so
[tex]x=0[/tex] ----> the y-axis
To graph the parabola, plot the given points and connect them
we have
The vertex is the point [tex](0,98)[/tex]
The y-intercept is the point [tex](0,98)[/tex]
The t-intercepts are [tex](-2.475,0), (2.475,0)[/tex]
The axis of symmetry is the y-axis
The graph in the attached figure
Part B) How far is the artifact fallen from the time t=0 to time t=1
we know that
For t=0
[tex]h(t)=-16(0)^{2}+98[/tex]
[tex]h(0)=98\ ft[/tex]
For t=1
[tex]h(t)=-16(1)^{2}+98[/tex]
[tex]h(1)=82\ ft[/tex]
Find the difference
[tex]98\ ft-82\ ft=16\ ft[/tex]
Part C) Does the artifact fall the same distance from time t=1 to time t=2 as it does from the time t=0 to time t=1?
we know that
For t=1
[tex]h(t)=-16(1)^{2}+98[/tex]
[tex]h(1)=82\ ft[/tex]
For t=2
[tex]h(t)=-16(2)^{2}+98[/tex]
[tex]h(2)=34\ ft[/tex]
Find the difference
[tex]82\ ft-34\ ft=48\ ft[/tex]
so
The artifact fall 48 feet from time t=1 to time t=2 and fall 16 feet from time t=0 to time t=1
therefore
The distance traveled from t=1 to t=2 is greater than the distance traveled from t=0 to t=1
