Can someone help explain this?

  • Thread starter RubyRed
  • Start date
  • Tags
    Explain
In summary, the object will be 15 meters above the ground after 4.9 seconds and it will strike the ground after t seconds.
  • #1
RubyRed
27
0
Hi there, I am in college algebra. I haven't done any of this in 3 years and am a bit rusty, I am very far behind and don't understand how to do this! Can someone please not just give me answers but help explain how to do a problem for me? The problem is this...

An object is propelled vertically upward with an initial velocity of 20 meters per second. The distance s(in meters) of the object from the ground after t seconds is s=4.9t^2+20t

a) When will the object be 15 meters above the ground?

b) when will it strike the ground?

c) Will the object reach a height of 100 meters.

I am guessing that with the a) you would make s=15 but as for what to do from there I have no idea. Same for be except making s=0?? And as far as c I don't have a clue. How could you answer c without knowing the rate that the object slows down to know if it would ever make it that high? I just don't get it!
 
Physics news on Phys.org
  • #2
the equation you have is slightly wrong... it should be s=-4.9t^2+20t and yes it would be a good idea to set s=15 m and solve for t (remember the quadratic equation?) same for b. as for c set it to 100 and see if you can get a real number (watch out for square roots of negative numers)
 
  • #3
sorry about the misprint! ok so

-4.9t^2 + 20t = 15 so then do I make it all = to 0. So subtact the 15 from both sides? Or do I divide everything by -4.9 to get the coefficient to the square be 1?
 
  • #4
start by subtracting 15 from both sides... the quadratic equation allows you to have an 'a' values for ax^2+bx+c, it is just usually equal to 1 in easier problems.
 
  • #5
ok so this is what I have done so far...

-15 + 20t - 4.9t^2=0

divided everything by -4.9

+15/4.9 - 20t/4.9 + t^2 = 0

subtract 15/4.9 from both sides to get -20t/4.9 + t^2 = -15/4.9

Am I on the right tract of way off? I am not sure what to do now. Do I add the 20t/4.9 to both sides to get the t^2 by itself. And then divde by t on both sides again to get t alone?
 
  • #6
remember the quadratic equation? maybe you forgot so here it is:
[tex]x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
you already have your problem set up in the form of [tex]ax^2+bx+c=0[/tex]
 
  • #7
oh boy...I think I am just doomed. lol I don't remember ANYTHING
 

Related to Can someone help explain this?

1. What do you mean by "explain this"?

"Explain this" is a request for further clarification or understanding about a particular topic or concept. It could refer to a specific idea, process, or phenomenon that the person is having difficulty comprehending.

2. Can you give an example to help explain this?

Sure! Giving an example is a great way to clarify a complex topic. For instance, if someone asks you to explain the concept of photosynthesis, you could provide an example of how plants use sunlight, water, and carbon dioxide to produce glucose and oxygen.

3. How do you explain something to someone who has no prior knowledge about it?

Explaining something to someone with no prior knowledge can be challenging, but it's important to break down the topic into simpler terms and avoid using technical jargon. You could also use analogies or real-life examples to help them understand.

4. What is the best way to explain a difficult concept?

The best way to explain a difficult concept will vary depending on the person you are explaining it to. Some people may prefer visual aids, such as diagrams or videos, while others may benefit from a step-by-step explanation. It's important to know your audience and adjust your explanation accordingly.

5. How can I improve my ability to explain things clearly?

Improving your ability to explain things clearly takes practice and patience. Some tips to help you include simplifying complex information, using relatable examples, and actively listening to your audience for any confusion or questions. You could also ask for feedback and continue learning about effective communication techniques.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
775
Replies
12
Views
965
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
187
  • Introductory Physics Homework Help
Replies
2
Views
559
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
809
Back
Top