A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
im new hi (moderators delete this please :) and my other question thanks)
anonymous
 one year ago
im new hi (moderators delete this please :) and my other question thanks)

This Question is Closed

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0well they tell you that v = 50 So what does the equation look like now?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0ok then they tell us that it starts off at 90 What do you get at the start of time? I.e. what do you get when \(t=0\)?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0nono, so if we let \(t=0\) into the function we get \(s\) correct?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0right, and they told us that the intial height is 90, so s=90. make sense?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0so what is the function?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0nono, ok before we can start to answer the question we had to figure out what \(v\) and \(s\) are. Now we have done that and we have our function is \[H(t) 16t^2+50t+90\] Are you with me?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0Now we can start to actually answer the question.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0Ok part A wants to know \(t\) when the thing hits the ground. What is the height when something hits the ground?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0ok so we need to solve \[16t^2+50t+90=0\] Can you do that?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0I am not sure if that is correct, but you can check that. Ok notice that since we are dealing with time, that we would only take the positive answer. That is part A. For part B, do you know how to find the vertex?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0the \(x\) coordinate of the vertex of the function \(ax^2+bx+c\) is \(x=\dfrac{b}{2a}\).

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0what is \(a\) and \(b\) in your equation? \(ax^2+bx+c\)?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0nono \(16x^2+50x+90\) \( \ \ \ ax^2+bx+c\) What is \(a\)?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0correct, ok so what is \(\dfrac{b}{2a}\)?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0correct, now plug that into the equation to get the height.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0I am not responsible for you failing a class. I also highly doubt that getting help on one question will change you from a failing grade to passing.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0so plug in 25/16 into the function and that will give you the max height.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.