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How do i compute distance if shot at 45degrees

  1. Apr 12, 2009 #1
    If I shoot a ball out of a cannon with muzzle velocity of 100 fps at 45 degree angle, how do I compute how far I shoot it? Ignore air resistance for this.
     
  2. jcsd
  3. Apr 12, 2009 #2

    mathman

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    Horizontal component of vel. is 100cos(45deg)=70.7, which stays constant. Vertical component is 100sin(45deg)=70.7, which is subject to gravity.

    The vertical distance as a function of time =vt-gt2/2, where v=70.7 and g=32. Set distance = 0 and solve for t. The non-zero solution will tell you the time it hit the ground, multiply by the horizontal velocity to get the distance.
     
  4. Apr 12, 2009 #3
    so for muzzle velocity Vm, I solve for t as indicated and get Vm/16=t.
    Then multiply that t by VmxCos(45) or Vm x.707; so it seems I simply
    multiply muzzle velocity by .707 to get the distance for any muzzle velocity.
    Is it that simple?
     
  5. Apr 13, 2009 #4

    Lok

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    I did not understand the last remark. But anyway:

    x=t*v*cos45

    t=(2*v*cos45)/g

    so x=(2*v2*cos452)/g where cos452=1/2

    => x=v2/g ( solution for distance in the case of 45' angle shooting)

    x-required distance
    v-muzzle velocity
    t-time it takes for the slug to rise and fall to earth

    Hope i did not do any mistakes :P...
     
  6. Apr 13, 2009 #5
    Shoudn't x=(2*v^2*t*cos45^2)
    and thus x=t*v^2/g instead of v^2/g
    Just curious....
     
  7. Apr 13, 2009 #6
    ?

    x = vxt

    Since the angle is 45 deg, and sin45 = cos45:

    vy = vx = v sin(45) = 100 sin(45)

    And t is the time for the object to go up and come down. So:

    y = vyt - (1/2)gt² = 0

    Could you solve it for t?
     
  8. Apr 13, 2009 #7
    Sure t=2Vy/a - thus 2x70.7/32=4.41875 secs
    which then gives 312ft for horizontal distance
    from 4.41875x70.7. This all from a muzzle velocity of 100 fps
    aimed at a 42degree angle

    is that correct?
     
  9. Apr 13, 2009 #8
    Very well :)

    Although I thought it was an angle of 45 deg. But you get the idea, so that's fine.
     
  10. Apr 14, 2009 #9

    Lok

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    No because

    x=t*v^2/g

    would mean that

    m=s*(m2*s2)*s2/m=m*s

    m is not equal to m*s

    basically the measuring units have to be the same.
     
  11. Apr 14, 2009 #10
    well...if you are shooting from level ground

    then the maximum range(R) of the projectile will be

    R= ( [tex]^{}u^2[/tex] )/g

    so your answer would be..approx 1020 feet...

    actually for a projectile fired from level ground on solving the various equations of motion you'll get

    Range = { [tex]^{}u^2[/tex]Sin2θ }/ g

    here θ is the angle of projection....

    this only works when the projectile is shot and lands at the same level...
    so it's best to stick to the euqations of motion...but in level to level cases, this formula can save time...
     
  12. Apr 14, 2009 #11
    What is u in your reply? thanks/Joe
     
  13. Apr 14, 2009 #12
    if horizontal flight time is t=2xVxcos(45)/a
    distance horizontally is txVxcos(45)
    combining them is 2xv^2*cos^2(45)/a and for
    100fps initial velocity on level ground, this means a distance of 312.5 feet
    Note that (from above) v^2sin(2x45)/g also works and gives 312,5 ft
    If this correct, thanks to all for all your help.
    if this is wrong, where am I wrong?
     
  14. Apr 14, 2009 #13
    Your answer is correct
     
  15. Apr 14, 2009 #14
    thanks to all for all your help
     
  16. Apr 14, 2009 #15
    sorry....forgot to mention this in my previous post,
    'u' is the initial speed of the projectile...
    in this case is would be 100fps
     
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