Question

CAN SOMEONE HELP ME PLEASE??? :-(
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 96 feet with an initial velocity of 80 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) = 31 + 32.2t 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?

Answer 2

Well for part A your initial velocity and height are given so you just need to plug them in.

Answer 3

H(t) = -16t^2 + 96t + 80?

Answer 4

Not quite. v = 80 so H(t) = -16t^2 + 80t + 96

Answer 5

oh ok

Answer 6

Can I assume you know calculus since looks like a Physics I problem, essentially?

Answer 7

Im taking algebra 1

Answer 8

Okay, give me a second.

Answer 9

Okay.

Answer 10

What you should be able to do is solve the equation for t when H(t) =96. This will give you the time required to launch and return to its launching height. Taking that time and cutting it in half will give you the time it takes to get to the maximum height. Plug that time back into the original equation to get the height it rises from the launch point. Does that make sense?

Answer 11

Kind of, so H(t) = -16(96)^2 + 80(96) + 96

Answer 12

No t is time, so basically, H(t) = -16t^2 + 80t + 96 = 96
From there, H(t) = -16t^2 + 80t = 0
Get the time, t = # then say max height time is t/2
Plug back into H(t/2) = -16t^2 + 80t + 96 gives you the height it rises.

Answer 13

Why 2? Because you cut it half?

Answer 14

Yeah, if it takes for example 10 seconds for the entire motion, half of that is rising, half is falling. We assume the midpoint is the max height (where it is neither falling nor rising).

Answer 15

so t=48?

Answer 16

Solving -16t^2 + 80t = 0 you get t = 48 seconds?

Answer 17

You should get two answers actually because its to the 2nd power. One will be impossible and the other is possible (the correct choice).

Answer 18

so i divided 96/2 which gave me 48 as t

Answer 19

no*

Answer 20

Ohh, no you need to solve -16t^2 + 80t = 0 for t. This will give you the time required. This is the algebra part.

Answer 21

Easiest way is to use quadratic formula. Have you learned that?

Answer 22

Yep that's the one.

Answer 23

-80+ or - sqroot of 80^2 -4(16)(0) over 2(16)

Answer 24

it should be -16 in each instance. The negative carries through but yes. That will give you the two answers.

Answer 25

\[\frac{-80\pm \sqrt{80^2-4(-16)(0)}}{2(-16)}\]

Answer 26

so -32?

Answer 27

@lasttccasey

Answer 28

No when you take the numerator you will get -80 +or- sqrt(80^2) which is going to give two answers. so -80 + 80 = 0 and -80 - 80 = -160. Now taking both numbers and dividing by 2(-16) you will get t = 0 and 5 seconds.

Answer 29

Does that make sense?

Answer 30

Okay now we know both times, since we know it is moving t = 0 is the impossible answer so t = 5 is the right one. Now we need half of this so t for the max height is 2.5 seconds.

Answer 31

From here we say H(2.5) to get the max height. Just plug 2.5 into each spot there is a t and it should give you the correct answers for part B

Answer 32

i got 16 @lasttccasey

Answer 33

Hmm, recheck your work. It should be H(2.5) = -(2.5)^2 + 80(2.5) + 96

Answer 34

what about the -16?

Answer 35

@lasttccasey

Answer 36

Yes sorry, it got deleted by accident.
H(2.5) = -16(2.5)^2 + 80(2.5) + 96

Answer 37

so 196? @lasttccasey

Answer 38

Yes, that is what I got.

Answer 39

Okay, thanks what about part c and d?

Answer 40

Part C: Another object moves in the air along the path of g(t) = 31 + 32.2t 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]

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

Answer 41

Okay so conceptually thinking what will happen if H(t) and g(t) are equal?

Answer 42

They would intersect?

Answer 43

Well yes if we can assume they are perfectly aligned but it really means they are at the same height from the ground. When you set the two equations together you will have a function of t. You will have to solve for t and this time is the time at which both objects are at the same height. Make sense?

Answer 44

Sort of

Answer 45

Well for example, if a person in California throws a ball exactly 3 feet in the air and a person in Florida drops a ball from 6 feet height. They will be at the same height at a specific time but they won't intersect necessarily.

Answer 46

oh

Answer 47

Hopefully that didn't confuse you worse lol but basically you have to say
-16t^2 + 80t + 96 = 31 + 32.2t
The question wants you to pick and chose random times and plug them in to the above equation until they are very similar so basically narrow down to an approximate one. Start from 1 and go to 15 seconds or something.

Answer 48

Okay, can you do it step by step?

Answer 49

Well is 12:38 AM where I'm at so I have to be going to bed but basically start at 1 second.
time, t = 1
-16(1)^2 + 80(1) + 96 = 31 + 32.2(1)
-16 + 80 + 96 = 31 + 32.2
160 = 63.2 (this is not close so try a different time)
t = 2
-16(2)^2 + 80(2) + 96 = 31 + 32.2(2)
-64 + 160 + 96 = 31 + 64.4
320 = 95.4 (still not close, I'm going to jump ahead because I solved for the exact answer)
... do this until you get close ...
t = 4
-16(4)^2 + 80(4) + 96 = 31 + 32.2(4)
-256 + 320 + 96 = 31 + 128.8
160 = 159.8 ( close enough)
The approximate answer is t = 4.0025
At this time the object and projectile are at the same height (assuming they are aligned they would intersect here)

Answer 50

For part H, once you get the approximated time, if it is over 2.5 seconds then the projectile is falling. From part B we saw the projectile stopped rising at 2.5 seconds.
Since t = 4 seconds basically then the projectile is falling when it collides.

Answer 51

Part D *

Answer 52

Oh okay, thank you very much for the help

Answer 53

Wait why 160?

Answer 54

Well 160 was just the numbers it came to be. -256 + 320 + 96 = 160 when doing this you have to treat both sides of the equation as separate so you can compare the results when you plug in the answer. You're welcome, I just hope you understood everything and I wish you luck in your class.

Answer 55

Oh ok, thanks