Question

The function H(t) = -16t2 + vt + s shows the height H (t), in feet, of a projectile launched vertically from s feet above the ground after t seconds. The initial speed of the projectile is v feet per second.

Part A: The projectile was launched from a height of 82 feet with an initial velocity of 60 feet per second. Create an equation to find the time taken by the projectile to fall on the ground. (2 points)

Part B: What is the maximum height that the projectile will reach? Show your work. (2 points)

Answer 1

Part C: Another object moves in the air along the path of g(t) = 10 + 63.8t where g(t) is the height, in feet, of the object from the ground at time t seconds.

Use a table to find the approximate solution to the equation H(t) = g(t), and explain what the solution represents in the context of the problem? [Use the function H(t) obtained in Part A, and estimate using integer values] (4 points)

Part D: Do H(t) and g(t) intersect when the projectile is going up or down, and how do you know? (2 points)

Answer 2

@jdoe0001 can you help me im lost on this

Answer 3

well. have you covered parabolas yet? or quadratic equations?

Answer 4

i know i little about parabolas but im not good at solving them

Answer 5

hmmm ok.... can you see the given equation, for the initial speed, is a parabola?

Answer 6

no not really

Answer 7

i see H(t) = -16t2 + vt + s but thats the height

Answer 8

well... have you covered quadratic equations yet?

Answer 9

yes

Answer 10

but this whole thing confuses me

Answer 11

well... a quadratic equation has the graph of a parabola

notice these 2 following ones
if we put a negative value in front of the leading term, or \(x^2\) it flips over

http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiJ4XjIrMngrMSIsImNvbG9yIjoiI0M3MkEyQSJ9LHsidHlwZSI6MCwiZXEiOiIteF4yKzJ4KzEiLCJjb2xvciI6IiMyODQ2REUifSx7InR5cGUiOjEwMDB9XQ--

Answer 12

now look at yours, it does have the a negative leading term
thus is a parabola opening downwards

Answer 13

it has a -16

Answer 14

\(\bf h = -16t^2+v_ot+h_o \qquad \text{in feet}\\

\\ \quad \\

v_o=\textit{initial velocity of the object}\\
h_o=\textit{initial height of the object}\\
h=\textit{height of the object at "t" seconds}\)

part A says, that the "initial height" is 82 and the "initial velocity" is 60
and using those values... .get an equation for the projectile trajectory

as you can see, the initial velocity formula already has those 2 factors, so part A is just
plug in those 2 values

Answer 15

so somthing like this f(t)= -16t2 + 60t + 82

Answer 16

is this correct or did i mess up

Answer 17

yes

Answer 18

how do i do part b?

Answer 19

recall that the parabola has a negative leading term, thus is opening downwards, thus so to find part B, or the maximum height

you'd just need to find the vertex of the parabola
you can always find the vertex of a parabolic equation at \(\bf \textit{vertex of a parabola}\\ \quad \\
y = {\color{red}{ -16}}x^2+{\color{blue}{ 60}}x+{\color{green}{ 82}}\qquad

\left(-\cfrac{{\color{blue}{ b}}}{2{\color{red}{ a}}}\quad ,\quad {\color{green}{ c}}-\cfrac{{\color{blue}{ b}}^2}{4{\color{red}{ a}}}\right)\)

Answer 20

how do i find the vertex can you help me with this real quick because i need to be able to show my work so i cant use a graphing program

Answer 21

well.. look above ^

Answer 22

ok so how do i simplify that to get the vertex

Answer 23

plug in the values and simplify the fractions

Answer 24

so it would equal -56.25

Answer 25

well... revise your simplification.... is 2 coordinates, one for x and one for y
and neither is negative

Answer 26

well.. one coordinate with 2 (x,y) values