anonymous
  • anonymous
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 :)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
welshfella
  • welshfella
horizontal component of the velocity = 1300 cos 45
welshfella
  • welshfella
equate this to d/5 to find d
anonymous
  • anonymous
So it would be 919.23 ?

Looking for something else?

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

More answers

anonymous
  • anonymous
Sorry I mean 183.84
welshfella
  • welshfella
that would be the horizontal velocity yes
welshfella
  • welshfella
919.23 = d / 5 find d
anonymous
  • anonymous
4596.15?
anonymous
  • anonymous
Thank you
welshfella
  • welshfella
looks good
welshfella
  • welshfella
yw
anonymous
  • anonymous
I got these out as I;m doing physics work right now too so thought I'd leave these here: \[V=V_0 + at\]\[X=1/2 (V_0+V)t\]\[X=V_0T+1/2(AT^2)\]\[V^2=V_0^2+2ax\]
welshfella
  • welshfella
the horizontal component of the velocity is constant as gravity has no effect on it
welshfella
  • welshfella
yes these are the equations for constant acceleration
welshfella
  • welshfella
- useful amongst other things, for solving problems of projectiles moving under the infuence of gravity

Looking for something else?

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