A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

YumYum247

  • one year ago

elp!!

  • This Question is Closed
  1. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hold up...

  2. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  3. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  4. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  5. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @IrishBoy123 Please check my work!!!! :)

  6. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    will look at it

  7. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    aight thanks :)

  8. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is my method of approaching the problem correct??????/

  9. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    absolutely 2 points 1 the question is unclear, are we assuming circle is centred at 42m or starts at 42m? in practise, this doesn't not matter, as the numbers are out by miles, but you should spot this when you are drawing the thing. perhaps you did. 2 yes, i agree with conclusions, but a/ your rounding is out for starters. 34 cos(18) = 32.33 b/ you used the quadratic formula when you had something simpler in your answer, you're using vector arrows and stuff that doesn't look right to me, so if we just call vertical displacement y(t) then you have \(y(t) = 10.5 t + \frac{1}{2}(-9.8)t^2 = t (10.5 - \frac{1}{2}(9.8)t)\) so \(y = 0 \implies t = 0 , \) or \(10.5 - \frac{1}{2}(9.8)t = 0\) so that solves to t = 2.14, fine, but I get 69.27, ie 69.3 for the distance it lands. small potatoes, but this is from the rounding. IOW fine but you could tighten it up a bit not sure what the third page is about.

  10. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    as for the waffly bit, i'd personally think a/ change elevation of cannon b/ change velocity of launch c/ change distance between cannon and targe you point about changing elevation to above 45 shows a good understanding and is well made other solutions, eg raising/lowering the cannon, making the target or the balloon massive, asking your deity to change the value of gravity, they all work too.....but not sure about that and i don't know your instructor yep, well done

  11. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but i actually went with manipulating the angle from 18deg( given) to a different range of angles from 78 to 70 to even 80 but the thing is that none of them quite land the balloon right on the target....i've about 1 to 2 meters short.... :"(

  12. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you mean you tried to find the exact angle?

  13. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes :D

  14. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    should i try to manipulate the vertical displacement of the launcher and keep the angle of the cannon the same 18deg...would that work?

  15. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but that would cause the balloon is gain more distance on the horizontal axis and my balloon will end up going way over my mark....42m :(

  16. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what du i du?

  17. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    do what you just did but be a bit more general if we use x and y for 'across' and 'up' then \(x(t) = 34 cos \theta \ \times \ t\) where \(\theta\) is our magic angle if it reaches \(x = 42\) at time \(t = T\), we can say \(T =\frac{42}{34 cos \ \theta} \) also we can say that \(y(T) = 0 \) because it has landed so \(0 = (34 sin \ \theta \ T - \frac{1}{2}gT^2) = 34 sin \ \theta - \frac{1}{2}gT = 0\) so \(34 sin \ \theta = \frac{1}{2}gT = \frac{1}{2}(9.8)\frac{42}{34 \ cos \ \theta}\) so \(2 sin \ \theta \ cos \ \theta = 2 \frac{1}{2}(9.8)\frac{42}{34^2 }\) the LHS is also \(sin \ 2 \theta\), this is a double angle identity that gives \(sin \ 2\theta = 0.356\) \(\theta = 10.43^o\) try that

  18. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    aight let me try that....:)

  19. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes it worked!!!! :"D Thanks and I Luv your foreverrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr!!!!!!!! :"D

  20. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    the *2* solutions, as you correctly identified there would be, are \(sin 2θ=0.356 \implies 2 \theta = 20.86^o, \ 180^o - 20.86^o \implies \theta = 10.43^o, \ 79.57^o \) there are rounding errors in there, now

  21. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.