The function 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 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. 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) = 20 + 38.7t 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?
@mathmale @triciaal @mathstudent55 @sleepyjess @ganeshie8 @zepdrix
HI!!
Hi
\[H(t) = -16t^2 + vt + s\] put \(v=60,h=100\) to get your function
that takes care of part A
do you know how to find the vertex?
\[h(t) = -16t^2 + 60t + 100?\]
yes that is it
Yes, -b/2a to find x coordinate. Input x coordinate in function to find y
i meant put \(x=100\) but you got it right
yes that is right
looks like you are in good shape
@misty1212 shouldn't s = 100?
You already have the equation: 0 = -16*t² + 80*t + 96; a quadratic in t. Use the quadratic formula to find t.
yes i said it wrong twice
Hold on, my internet is freaking out
\(s=100\) your function is correct
@boldjon Do I need to find t for any of the parts? i have no idea lol
the function @boldjon has must be from some other question, not this one
\[x = \frac{ -b \pm \sqrt(b^2 - 4ac) }{ 2 }\]
wait, that's wrong
yes it is, but you do not need that, that is for where \(h=0\) i.e when it hits the ground, which i don't think you are asked
\[x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a }\]
alright
So Part A is done as we have already created the equation. Part B i can probably solve on my own as I know it's -b/2a and then input the x coordinate into the function
I'll do part b anyways though Part B: What is the maximum height that the projectile will reach? Show your work
H(t) = -16t^2 + 60t + 100 I believe this is written in ax^2 + bx + c form, correct @misty1212 ?
eys
alright, so a = 16, b = 60, and c = 100
no \(a=-16\)
oh right
-b/2a = -60/2*-16 2 * -16 = -32 -60 / -32 = 1.87500 Which can be rounded down to 1.88
i would leave it as a fraction, reduce it, not convert to a decimal
alright
\[\frac{15}{8}\]
or go ahead and use \(1.875\) if you like, you will probably be best off with a calculator anyways
I work better with decimals, can we do the exact decimal form instead?
alright, thanks
yeah go ahead and use \(1.875\)
i can check what you get if you would like
H(1.875) = -16(1.875)^2 + 60(1.875) + 100 Solve as usual 1.875^2 = 3.515625 3.515625 * -16 = -56.25. 60 * 1.875 = 112.5 Add together -56.25 + 112.5 + 100 = 156.25
yup easy if you use this http://www.wolframalpha.com/input/?i=-16%281.875%29%5E2%2B60*1.875%2B100
That's strange, I input the equation into desmos and it says it's (-1.88, 43.75) @misty1212 any ideas?
no but you and wolfram are right probably something with parentheses
oh you should use \(1.875\) not \(-1.88\) 'the sign is wrong there
Now it's correct. For some reason desmos didn't understand h(t), probably a bug. I switched all the t's with x's and now it says the correct vertex
(1.88, 156.25)
@misty1212 That solves Part B Now Part C 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 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]
I think we can use the draw tool for this
i guess it wants you to check with whole numbers
i would just go ahead and set them equal and solve if you can use the computer, do this http://www.wolframalpha.com/input/?i=-16x%5E2%2B60x%2B100%3D20%2B38.7x
Right, I think I can do Part C in a word docuement anyways
Part D: Do H(t) and g(t) intersect when the projectile is going up or down, and how do you know?
did you find when they intersect? you need that first
2 places when x is abotu -1.66 and when x is about 2.99
they want you to use whole numbers here the first one does not make any sense, it was in the past 2.99 rounds to 3 they meet in 3 seconds
as for going up or going down, is 3 seconds before or after the projectile reaches its maximum height?
I believe after, correct?
yes so it is going down
Alright, thanks!
@misty1212 I am having trouble. Does g(t) even have a maximum height?
@misty1212 Never mind, I'm good. Thanks for all your help on this!
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