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.
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) = 31 + 32.2t where g(t) is the height, in fe

- anonymous

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- anonymous

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?

- anonymous

@perl @perl

- perl

Which part would you like help with?

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- anonymous

all of it.....

- perl

Ok we can start with part a)

- anonymous

maybe once i get started i MIGHT be able to do it
thanks so much

- anonymous

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

- anonymous

i think ^^

- perl

We are given that \( H(t) = -16t^2 + vt + s\) 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.
That means s = 96 and v = 80

- perl

yes that's correct :)

- anonymous

ok so my equation becomes : H(t) = -16t^2 + 80t + 96

- anonymous

ok so thats part A right?

- anonymous

now part b i have to find the maximum

- anonymous

so the axis of symmetry is 2.5?

- perl

the maximum or minimum of a parabola occurs at \(\large t= \frac { -b}{2a} \)

- anonymous

ok so I got a maximum of 196

- perl

how plug in that time t = 2.5 to find the maximum height

- anonymous

196 :)

- perl

correct

- anonymous

ok so now part c... now i really dont know where to start

- perl

Part C) the directions are not complete . Can you check

- anonymous

yes hang on one sec please

- anonymous

Part C: 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.

- anonymous

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]

- perl

We want to find when \( H(t) = g(t) \) which would ordinarily mean solve the equation \( -16t^2 + 80t + 96= 31 + 32.2t \)
but instead of solving for t , here they ask us to make tables for the left side and right side separately, and estimate when they are closest

- perl

|dw:1435527312454:dw|

- anonymous

ok so I would solve each equation both H(t) and g(t) by plugging in the numbers 1-5?

- perl

we can find y values for both H(t) and g(t) and then estimate where they are closest

- anonymous

ok so we would do that by plugging in 0 right?

- perl

yes

- anonymous

ok so g(t) = 31 + 32.2t , g(t) = 31 + 32.2(0) , g(t) = 31 + 0 , g(t) = 31
H(t) = -16t^2 + 80t + 96 , H(t) = -16(0)^2 + 80(0) + 96 , H(t) = 0 + 0 + 96 , H(t) = 96

- anonymous

so the y-intercept for H(t) = -16t^2 + 80t + 96 is 96 and the y-intercept for g(t) = 31 + 32.2t is 31

- anonymous

@perl

- anonymous

can you help me with part d @perl

- anonymous

equation 1: (4,160)
equation 2: (4, 159.8)
what does it represent in the context of the problem?

- perl

|dw:1435528016137:dw|

- perl

|dw:1435528443369:dw|

- anonymous

they almost intersect when x = 4
what is the meaning in the context of the problem?

- anonymous

@perl

- perl

at time t = 4 seconds the paths are closest to each other

- perl

the projectiles are closest

- anonymous

ok thanks! what about part d?

- anonymous

when the projectiles re going down because H(t) = -16t^2 +80t +96 is decreasing just as the other equation is increasing?

- anonymous

@perl

- anonymous

@perl is that correct??

- perl

when the projectile h(t) is coming down , yes

- perl

the projectile g(t) is going up the whole time

- anonymous

ok thanks so much !!!!!!

- perl

|dw:1435528915392:dw|

- perl

rough picture :)

- anonymous

THANK YOU SO MUCH!

- perl

Your welcome

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