A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

idku

  • one year ago

I need help with this (math-based-physics)

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

    A cannonball is shot (from ground level) with an initial horizontal velocity of 33 m/s and an initial vertical velocity of 25 m/s. ------------------------------------------------------------ a) What is the speed of the cannonball 3.7 seconds after it was shot? b) How high above the ground is the cannonball 3.7 seconds after it is shot?

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

    I found so far: [1] initial speed, \(\rm u=41.4(m/s)\) [2] angle of cannonball with respect to ground, \(\rm \theta=37.15º\) [3] maximum height the cannonball goes above the ground \(\rm x=31.8(m)\) [4] horizontal displacement is \(\rm 167.6(m)\) [5] the entire time of the ball motion \(\rm 5.08(s)\) ------------------------------------------------- I need: [6] speed of the cannonball at, \(\rm t=3.7(s)\) [7] height (vertical displacement) at, \(\rm t=3.7(s)\)

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

    @jim_thompson5910 can you help me please when you are free?

  4. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I agree with your answers to [1] and [2]. I got the same results. I'm now checking [3]

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

    Oh, no need to check. I am entering answers in my pracitce tool. and I got how to do 1-5 all.

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

    I have managed myself through 1-5 with a great pain, nd now trying to understand why V=u+at doesn;t work for me. So I am using this: V=u+at {1} t is time {2} a is acceleration due to gravity. {3} where u is the initial velocity -> (vertic. veloc., because for horiz. a=0) V=25(m/s) + (-9.81 m/s\(^2\))(3.7s)=-11.297m/s (and that is wrong)

  7. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm not the best with physics, but I'm guessing you can find the position function and then take the derivative to find the velocity function. After that, plug in t = 3.7 to find the speed at time 3.7 seconds.

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

    so if we have gone from 25m/s to 0 m/s how many second would that have took if every second we lose 9.81m/s that is 25/9.81 seconds= 2.55 so that was 1/2 from start to vertex. Now from vertex to end, I get 5.1 better answer for time

  9. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    also, this page might help http://formulas.tutorvista.com/physics/projectile-motion-formula.html it's helped me get refreshed on the projectile formulas

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

    Oh, I am also kind of like that. I am good at math, but my physics skills are like a fahrenheit temperature in artarctia during winter season (July-August)

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

    Oh, that looks like it can help. tnx will see.

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

    I was chatting with a tutor and they asked bucks....

  13. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh don't chat there lol that's a bot or a sales rep

  14. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    just close that portion out and focus on the formulas on the page

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

    From previous posts I had a formula \(V=u+at\) it is correct. (Velocity is integral of acceleration w/ respect to time, + initial value of our initial velocity u) So, velocity at at t=3.6 should be given by V=25m/s -9.81(m/s\(^2\))•[5.1(s)-3.7(s)]=11.266m/s but that is incorrect.

  16. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you're only considering the vertical component though

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

    so instead of 25m/s I should take 41m/s /

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

    ?

  19. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well I'm thinking it would be the result of [1]

  20. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    since that's the composition of the vertical and horizontal speeds

  21. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm not sure though

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

    Maybe I can tag @dan815

  23. 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.