simple math

[Ron Coleman]

Lets say that I loaded a 14 lb. round ball in my recent project and fired it at a high angle. Lets say that I found the ball 2400' away and that my video showed that the hang time of the ball was 20 seconds. How high would the ball have gone?
How did you figure it out?
How much FG do you think that would require?



RonC

[StrawHat]

Pretty sure that the simple math you request is in a simple physics class.

The one I didn't do so well in.

I'd look it up but the physics book is propping up my coffee pot table.

However, there is hope.

http://www.go2gbo.com/forums/index.php?board=88.0

They might be able to help you.

Let us know what you discover.

[garandsrus]

Ron,

I am sure someone will correct this if it's wrong, but in terms of "simple math" the only thing that matters as far as the max height is concerned is the hang time. The formula should be:

Height = 1/2*acceleration due to gravity*time*time.

You would need to make an assumption that the time up and down is the same, which it may not be due to the initial velocity of the cannon ball on the way up verses the terminal velocity of the cannon ball on the way down.

If you make the assumption above, the formula for 1/2 the ball travel (up or down, 10 seconds instead of 20) would be would be:
Height = (.5)(32)(10)(10) = 1,600 ft.

A 10 second hand time (total trip) would be (.5)(32)(5)(5) = 400 ft. Quite a difference due to the acceleration over time.

John

[Nueces]

Hi, Ron, welcome aboard. Here’s what I did in answer to your question.

John, I get the same result as you for the max height in a vacuum (no air resistance). The ball was also found 2400’ away from the muzzle. Assuming it landed there 20 seconds after the shot, that implies a horizontal velocity component of 120 feet/second. The vertical component, which yielded the 1600 feet max height, would have been 320 feet/second. These components resolve into a total muzzle velocity of 341.76 feet/second and a muzzle elevation of 69.44 degrees above the horizontal (if the shot was over level ground).

The equations used for the vertical part are:

height = MV * t - ½ * g * t**2

velocity = MV – g * t

where, MV is the vertical muzzle velocity, t is time in seconds, g is gravitational acceleration (32 feet per second per second), * denotes multiplication and ** denotes that t is raised to the power of two, or squared. The gravitational terms are subtracted because the initial velocity is up and g acts down. Ground, or muzzle, level is at zero height and the clock starts at the shot. Max height is achieved when the velocity at the top is zero.

Total MV and the angle are found by setting up a right triangle, with the short sides equal to the vertical and horizontal components and using the Pythagorean theorem to get the hypotenuse for total MV. The angle is the inverse tangent of 320/120, or 69.44 degrees.

Mark

It’s late and I’m tired, so this may look different in the a.m., but I give it about a 78% chance of being OK. :mrgreen:

[Ron Coleman]

Thank you gents, I'm 100% happy with your answers. Bright blue bowling balls are highly visible, and GPS units are handy.

RonC