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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) 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)
YEAH, I have that part. but what do you mean by A;0 .. I have the numbers plugged in .. just solving is difficult
A: 0. Sry. Forgot the space
Just sayin for problem a.
either way what do you mean??
The equation for problem a is 0=-16t2+60t+82
I know that
part A is definitely 0 = -16t^2+60t+82. So you have the correct answer there for that part. initial velocity = 60 ft/s v = 60 initial height = 82 ft s = 82
for part b, consider the equation y = -16x^2+60x+82 it is in the form y = ax^2 + bx + c
okay... now what do i do?
notice how a = -16, b = 60, c = 82 plug the values of a and b into h = -b/(2a) and tell me what you get for h
it should be positive
you lost a sign somewhere
yea. it is . i accidentally put it there
now plug x = 1.875 into y = -16x^2+60x+82 to get the value of y what is the value of y?
that's too big
,,,,, but i plugged the number in and I'm sure i solved it correctly..
-16x^2+60x+82 -16(1.875)^2+60(1.875)+82 ... replace every x with 1.875 now compute `-16(1.875)^2+60(1.875)+82`
yeah.. thats what i had. .let me retry
okay great.. next step?
For c, u make the equation in a equal 10 +63.8t 10+63.8t=-16t2+60t+82 In which you solve for t, then plug it in in one of the equations
So when the time is t = 1.875 seconds, the height of the object is 138.25 ft which is the peak height |dw:1439942710928:dw|
I thought u were done with b, sry for interupting😰
what kind of calculator do you have?
So basically, 138.25 meters is the answer to b
Texas Instruments TI 30X
have you used a website called desmos before? It's a free graphing calculator
ok we're going to use that to answer part c
ok what do i type in?
type in f(x) = -16x^2 + 60x + 82 into the first box and g(x) = 10 + 63.8x into the second box let me know when you have that in and I'll move onto the next step
You want go and fx to be the same
Gx dang it autocorrect
okay done.. but i don't know how to put the f(x) and g(x) into the equation.. I don't see those as options.. like i can type the like equation after the = sign but nothing before
how did you get the f(x) there... just type?
ok.. but no lines appear
If Gx=Fx, the -16t2+60t+82= 10+63.8t
your graph window may be off
i have a mac.. nothing is wrong with it.. so how is that possible?
Des ps is a lot quicker :3
try setting the y min to -20 and the y max to 150
click the wrench on the right side to adjust the graph window
i have lines now..
now onto the table part select the box under the g(x). So select the third box then move up to the + button and add a table change the \(\Large x_1\) to just x. In the second column, replace \(\Large y_1\) with f(x). You'll see a list of numbers pop up in the third column, type g(x) at the top of this column. More numbers will pop up
the goal is to have the numbers be equal. Which row does this? Or do we get pretty close at all?
they aren't appearing with numbers. i have the table and stuff.. but there aren't any numbers
okay I got it now
do you see any rows where f(x) = g(x) or pretty close to it?
yea. row 2
yep, so at about t = 2 seconds, the two objects will be at the same height (approximately)
awesome! So is that part c??
okay and Finally, Part D/
and it might be hard to see, but the projectile is falling down when they finally met up here is a zoomed in look |dw:1439944300088:dw|
if you click the intersection point between the two graphs, you'll see the point (2.006, 137.98) pop up. This is the more accurate approximate intersection point So at approx t = 2.006 seconds, the two objects are at an approx height of 137.98 ft
@jim_thompson5910 so if that is all.. I know your probably exhausted ...but could you help me with another problem I only need help on one part and its fairly simple???