Question

Helppp.

Answer 1

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 90 feet with an initial velocity of 50 feet per second. Create an equation to find the time taken by the projectile to fall on the ground.

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

Part C: Another object moves in the air along the path of g(t) = 28 + 48.8t 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]

Answer 2

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

Answer 3

I just need help with C and D

Answer 4

h(t) = -16t^2 + vt + s
h(t) = -16t^2 + (80)t + (96)
h(t) = -16t^2 + 80t + 96

Answer 5

what one is that..?

Answer 6

@Diamond_Swag11

Answer 7

that is step zero, before you can do part A

Answer 8

it is also wrong

Answer 9

oh. I have already completed part a and b

Answer 10

looks like @Diamond_Swag11 made the numbers up, but you were given the numbers in the problem

Answer 11

\[h(t) = -16t^2 + 50t + 90\] is your equation
what did you get for the max height?

Answer 12

H(t)=−16t2+50t+90 the maximum height occurs at the vertex

so the maximum height = 2065/16, occurs at t = 25/16

Answer 13

ok good

Answer 14

part c is to solve
\[ -16t^2 + 50t + 90=48.8t+28\]

Answer 15

it says to use a table of values, but we can do it on the computer

Answer 16

http://www.wolframalpha.com/input/?i=-16t^2+%2B+50t+%2B+90%3D48.8t%2B28
looks like the solution is very very close to \(t=2\)

Answer 17

you good from there?

Answer 18

that was part c correct?

Answer 19

yes, that was part C

Answer 20

Can you also help with part D? thats the last thing I need help with!

Answer 21

yes but you can do it too
the first one had a max height when \(t=\frac{25}{16}\)
before that it was going up
after that it is going down
is 2 bigger than or less that \(\frac{25}{16}\)?

Answer 22

bigger

Answer 23

then it was going down right?

Answer 24

yes

Answer 25

the end

Answer 26

lol. Thank you!!<3(: