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)

Question

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)

anonymous · Answer

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?

anonymous · Answer

medal &fan!

kirbykirby · Answer

A) 
You have that $-16t^2 + vt + s$
According to what you are given, $v=60$ and $s=82$
so $H(t)=-16t^2+60t+82$ 

Now you know that the projectile will hit the ground at height 0, so $H(t)=0$, so
$0 = -16t^2+60t+82$ then you can solve for $t$ by using the quadratic equation

anonymous · Answer

I got  h(t)=-16^2+80t+96 lol how did I mess up?

kirbykirby · Answer

where do 80 and 96 come from?

anonymous · Answer

..oh, I was looking at a different problem -__- Sorry bout that

kirbykirby · Answer

For B).. notice that you have a parabola of the form $at^2+bt+c$ , where $a=-16$. It's negative, so the parabola will have this shape: $\cap$

You know then that the maximum will occur at the vertex, so one thing you can do is transform your equation into vertex form: $a(x-h)^2+k$ where the vertex has coordinates $(h, k)$ and is the maximum.

$$-16t^2+60t+82\
=-16\left( t^2+\frac{60}{-16}t+\frac{82}{-16}\right) $$ Add and subtract $\left(\dfrac{b}{2}\right)^2$ to complete the square: $\left(\dfrac{\frac{60}{-16}}{2}\right)^2=\dfrac{225}{64}$ 

$$-16\left( t^2+\frac{60}{-16}t+\frac{82}{-16}\right) \= -16\left( t^2+-\frac{15}{4}t+\frac{225}{64} - \frac{225}{64}-\frac{82}{16} \right) \
=-16 \left( t-\frac{15}{8}\right)^2 -16\left( -\frac{553}{64}\right) \
=-16\left(t-\frac{15}{8}\right)^2\+\frac{553}{4}$$ 

So the vertex is at $\left(\dfrac{15}{8},\dfrac{553}{4}\right)$, so the max height is 553/4 at t=15/8

kirbykirby · Answer

I'll let you absorb that. If anything is not clear, let me know before I continue

anonymous · Answer

sorry, yeah I get it for B

kirbykirby · Answer

For C).. I;m not sure why they want you to approximate it when you can just find the exact solution using the quadratic formula.. 

$g(t)=H(t)\
10 + 63.8t=-16t^2+60t+82\
0=-16t^2+60t+82-10-63.8t\
0=-16t^2-3.8t+72$

$$t=\frac{-(-3.8)\pm\sqrt{(-3.8)^2-4(-16)(72)}}{2(-16)}\
t=\frac{3.8\pm\sqrt{4622.44}}{-32}\
t= -2.2434  ~ or~ t=2.0059$$

I suppose if you need to approximate with an integer using a table... Then you can just plug in integer values for t, like t=1, t=2, t=3, t=4, etc. And you should see that t=2 gives the closest true approximation. (And from above, we see that it is true since t=2.0059 is the actual answer). We usually ignore the negative sign because in this physical context, $t$ represents time, and we don't have negative values for time. 

And you find here represents the amount of time your projectile took to hit the other object. If you graph it, it would be something like this: |dw:1411426265458:dw|

anonymous · Answer

for part C i don't see a question though, it just has information (i thought)

kirbykirby · Answer

Well they want you to find the solution of $g(t)=H(t)$. That is basically the question. I mean it's not in a "regular" question form with What/How and " ? ", but ya, that is what they ask for :)

kirbykirby · Answer

And they want the explanation of what it means

anonymous · Answer

Oh, okay i get it now lol

kirbykirby · Answer

For D... your parabola I drew it like this with this shape $\cap$ . So, the projectile is basically travelling from left to right: |dw:1411426728585:dw| 
So.. do you notice how the projectile is going "up" before reaching the maximum, and then going "down" after reaching the maximum? The question basically wants to know did the collision occur when the projectile was going up, o when it was going down (in other words,, before, or after the maximum?) To illustrate this: |dw:1411426857685:dw|