A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

A bullet is fired into the air with an initial velocity of 1,300 feet per second at an angle for 45° from the horizontal. What is the horizontal distance traveled by the bullet in 5 seconds? Can someone tell me what formula I should me using? I think I can do it myself but I can't remember velocity. Medals for everyone :)

  • This Question is Closed
  1. welshfella
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    horizontal component of the velocity = 1300 cos 45

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

    equate this to d/5 to find d

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

    So it would be 919.23 ?

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

    Sorry I mean 183.84

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

    that would be the horizontal velocity yes

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

    919.23 = d / 5 find d

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

    4596.15?

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

    Thank you

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

    looks good

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

    yw

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

    I got these out as I;m doing physics work right now too so thought I'd leave these here: \[V=V_0 + at\]\[X=1/2 (V_0+V)t\]\[X=V_0T+1/2(AT^2)\]\[V^2=V_0^2+2ax\]

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

    the horizontal component of the velocity is constant as gravity has no effect on it

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

    yes these are the equations for constant acceleration

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

    - useful amongst other things, for solving problems of projectiles moving under the infuence of gravity

  15. 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.