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## Main Question or Discussion Point

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.

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- #1

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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

mathman

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The vertical distance as a function of time =vt-gt

- #3

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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?

- #4

- 555

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x=t*v*cos45

t=(2*v*cos45)/g

so x=(2*v

=> x=v

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...

- #5

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Shoudn't x=(2*v^2*t*cos45^2)

and thus x=t*v^2/g instead of v^2/g

Just curious....

and thus x=t*v^2/g instead of v^2/g

Just curious....

- #6

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?Shoudn't x=(2*v^2*t*cos45^2)

and thus x=t*v^2/g instead of v^2/g

Just curious....

x = v

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

v

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

y = v

Could you solve it for t?

- #7

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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?

- #8

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Very well :)

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?

Although I thought it was an angle of 45 deg. But you get the idea, so that's fine.

- #9

- 555

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No becauseShoudn't x=(2*v^2*t*cos45^2)

and thus x=t*v^2/g instead of v^2/g

Just curious....

x=t*v^2/g

would mean that

m=s*(m

m is not equal to m*s

basically the measuring units have to be the same.

- #10

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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...

- #11

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What is u in your reply? thanks/Joe

- #12

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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?

- #13

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Your answer is correct

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

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thanks to all for all your help

- #15

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'u' is the initial speed of the projectile...

in this case is would be 100fps

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