anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • 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
  • anonymous
@perl @perl
perl
  • perl
Which part would you like help with?

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anonymous
  • anonymous
all of it.....
perl
  • perl
Ok we can start with part a)
anonymous
  • anonymous
maybe once i get started i MIGHT be able to do it thanks so much
anonymous
  • anonymous
H(t) = -16t^2 + 80t + 96
anonymous
  • anonymous
i think ^^
perl
  • 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
  • perl
yes that's correct :)
anonymous
  • anonymous
ok so my equation becomes : H(t) = -16t^2 + 80t + 96
anonymous
  • anonymous
ok so thats part A right?
anonymous
  • anonymous
now part b i have to find the maximum
anonymous
  • anonymous
so the axis of symmetry is 2.5?
perl
  • perl
the maximum or minimum of a parabola occurs at \(\large t= \frac { -b}{2a} \)
anonymous
  • anonymous
ok so I got a maximum of 196
perl
  • perl
how plug in that time t = 2.5 to find the maximum height
anonymous
  • anonymous
196 :)
perl
  • perl
correct
anonymous
  • anonymous
ok so now part c... now i really dont know where to start
perl
  • perl
Part C) the directions are not complete . Can you check
anonymous
  • anonymous
yes hang on one sec please
anonymous
  • 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
  • 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
  • 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
  • perl
|dw:1435527312454:dw|
anonymous
  • anonymous
ok so I would solve each equation both H(t) and g(t) by plugging in the numbers 1-5?
perl
  • perl
we can find y values for both H(t) and g(t) and then estimate where they are closest
anonymous
  • anonymous
ok so we would do that by plugging in 0 right?
perl
  • perl
yes
anonymous
  • 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
  • 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
  • anonymous
@perl
anonymous
  • anonymous
can you help me with part d @perl
anonymous
  • anonymous
equation 1: (4,160) equation 2: (4, 159.8) what does it represent in the context of the problem?
perl
  • perl
|dw:1435528016137:dw|
perl
  • perl
|dw:1435528443369:dw|
anonymous
  • anonymous
they almost intersect when x = 4 what is the meaning in the context of the problem?
anonymous
  • anonymous
@perl
perl
  • perl
at time t = 4 seconds the paths are closest to each other
perl
  • perl
the projectiles are closest
anonymous
  • anonymous
ok thanks! what about part d?
anonymous
  • anonymous
when the projectiles re going down because H(t) = -16t^2 +80t +96 is decreasing just as the other equation is increasing?
anonymous
  • anonymous
@perl
anonymous
  • anonymous
@perl is that correct??
perl
  • perl
when the projectile h(t) is coming down , yes
perl
  • perl
the projectile g(t) is going up the whole time
anonymous
  • anonymous
ok thanks so much !!!!!!
perl
  • perl
|dw:1435528915392:dw|
perl
  • perl
rough picture :)
anonymous
  • anonymous
THANK YOU SO MUCH!
perl
  • perl
Your welcome

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