Question

please someone help me :( I am so frustrated with these questions!!!!



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.

Answer 1

Part A: The projectile was launched from a height of 100 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)

Part C: Another object moves in the air along the path of g(t) = 20+38.7t where g(t) is the height in feet of the object form 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

Ok basically we are given an equation that describes the path of a projectile and part A gives us information to complete this equation

H(t)=-16t^2+vt+s
To start off H(t) is the height of the projectile at time t in seconds
s represents the initial height of the projectile and v represents the initial speed

Now in part A it tells us that the initial height of the projectile is a 100
and the initial speed is 60

Using this information and plugging in the values what would our equation be?

Answer 3

H(100)-16t^2+vt+60?

Answer 4

Nope its
H(t)=-16t^2+vt+s
Remember t represents time
v represents the initial speed and s represents the initial height

Answer 5

so it would be H(t)-16t^2+60t+100?

Answer 6

Yup so Part A is
H(t)=-16t^2+60t+100

Answer 7

okay cool part b now?

Answer 8

Part B is asking for the maximum Height
Now the maximum height occurs at the vertex. The x-coordinate represents the time that the maximum height occurs and the y-coordinate represents the maximum height
So we know a formula to find the x-coordinate of the vertex
\[x=-\frac{b}{2a}\]
Our equation is in the form y=a*t^2+b*t+c
So a=-16 , b=60 and c=100
So plugging in these values we get
\[x=-\frac{60}{2(-16)}=1.875\]
So our x-coordinate of the vertex is 1.875

Now to find the y-coordinate we plug in t=1.875 to find out the height

\[H(1.875)=-16(1.875)^2+60(1.875)+100=268.75

So the maximum height occurs at 268.75 ft

Answer 9

Do you follow so far?

Answer 10

someone told me this is the answer so im a little confused
Part B :

-16t^2 + 60t + 100

-b/2a = -60/2*-16 = 15/8

f(-b/2a) = -16(15/8)^2 + 60(15/8) + 100
= -16*225/64 + 15*15/4 + 100
= 625/4

Answer 11

Well thats the same thing
Compute on your calculator 15/8 and 625/4 and tell me what u get

Answer 12

1.875 and 156.25

Answer 13

Ohhhh I messed up. I forgot to include the negative sign
\[H(1.875)=-16(1.875)^2+60(1.875)+100=156.25\]

Answer 14

its okay can we do part c now?

Answer 15

Ok so basically we wanna know when h(t)=g(t)
So lets set the 2 equation to equal each other and then solve for t
\[-16t^2+60t+100=20+38.7t\]
\[-16t^2+60t-38.7t-20=0\]
\[-16t^2+21.3t-20=0\]

Now we gotta solve for t using the b formula.
You familiar with that?

Answer 16

Part C:

-16t2 + 60t + 100 = 20+38.7t

-16t^2 + 21.3t + 80 = 0
t = -2 or 3
Since the solution is a common point on both graphs, the solution t = 3 represents the time when the both objects cross each other.

Answer 17

would that be that answer i should put down for part c?

Answer 18

ohh yaaa I forgot abt the 100
So the equation would be -16t^2+21.3t+80=0
but ya in any case you case in order to get the answer you need to use the b formula. I wldnt be able to find t=-2 or t=3 just by looking at the equation. You need to use the b formula
And btw time cant be a negative number so the only answer would be t=3

Answer 19

and this is the answer for part d?
Part D :
H(t) and g(t) must intersect when the projectile is going down because the maximum height occurred at -b/2a = 15/8 which is less than the intersection point of t = 3. So, the both graphs must be intersecting when the projectile is fallind down.

Answer 20

Thats the correct answer

Answer 21

cool thanks for all your help :)

Answer 22

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] (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)