Question

Please help me! Last question!! Fan and Medal!

Answer 1

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 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. (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) = 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] (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)

Answer 2

I've already completed A

Answer 3

Hello again. Too bad we haven't yet finished the last two problems you posted.
In the present question, are you taking an ALGEBRAIC or a CALCULUS approach?

Answer 4

Please type out your equation for Part A, so that I can see how you've approached this problem.

Answer 5

I winged the last two problems because I have been in the same 5 question exam for coming up on 7 hours. I'm tired of it. I just need to put something logical for the answers, even if its not completely accurate, to get it submitted in the next 10 minutes.

Answer 6

A. h(t)=-16t^2+vt+s
h(t)=-16t^2+(80)t+(96)
h(t)=-16t^2+80t+96

Answer 7

Want to go back to them, or want to focus on the current problem set?

Answer 8

Current problem please. I have 10 minutes

Answer 9

Wouldn't that be h(t)=-16t^2+60t+100?

Answer 10

One of the faster ways of answering Part B would be to sketch the graph of h(t)=-16t^2+60t+100. You could "eyeball" the t value at which the ball reaches its max ht.

Algebraically, use the vertex formula t=-b/(2a) to determine the t coordinate of the point representing the max ht.

Answer 11

If you happen to know Calculus, take the derivative of H and set it equal to zero; then solve for t.

Which of these approaches would be most comfortable for you?

Answer 12

So by graphing it, I can see that the max height is about 156

Answer 13

Sounds reasonable. And at what time does that max ht occur? Draw a line straight down to the t axis from the point repr. the max. ht.

Answer 14

I estimate this time to lie between 1 and 2 secs.; What about you?

Answer 15

It happens at 5 seconds, roughly.

Answer 16

Unfortunately, there's a big discrepancy there.

Look at h(t)=-16t^2+60t+100.

What are the values of the coeff. a, b and c?

Answer 17

Once you have them, calculate t=-b/(2a).

Answer 18

that would give you the time coordinate of the vertex of this inverted parabola.

Answer 19

Review: x=-b/(2a) gives us the x coordinate of the vertex of the graph of y=ax^2+bx+c.

Here, you want t=-b/(2a) for h(t)=-16t^2+60t+100. Here a=-16, b=60. Please find
t=-b/(2a).

Answer 20

The question isn't asking me for time, only max height. I'm just going to move on to pt C because i'm so fed up with this exam. Can you help me with part C?

Answer 21

Sure. The idea is: in B you find the time at which the projectile reaches max ht, and once you have that time (1 7/8 sec), you substitute it into the equation -16t^2+60t+100 to obtain the max ht. So I was indeed focusing on helping you to solve this problem.

Answer 22

Now I'm going to look at Part C.

Answer 23

Sorry...typed a lot, then lost it when OpenStudy went AWOL for a few secs.

Answer 24

Basically, y ou are asked to equate h(t) and g(t).

In other words, you equate -16t^2+60t+100 to 20+38.7t and try to solve that for t.
Can you do that?
Doing so will be much like the previous problem (which we abandoned). OR ... better ... you could simply graph these 2 equations and estimate the time at which the graphs cross one another.

Answer 25

If you have a graphing calc., by all means use it.

Answer 26

They wont intersect because the second equation goes up to 256

Answer 27

Have you a graphing calc? Again, if yes, please use it. sorry to be pushing your idea aside, but I don't thing your 'goes up to 256' statement is relevant here.

Answer 28

The equation -16t^2+60t+100=20+38.7t does have a solution; the 2 curves do intersect; they intersect at approx. t=3.

Answer 29

Again, OpenStudy was down and we lost valuable time because of that.