- #1
Stobbe
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Ok, don't ask why I was doing this dynamics problem from an old textbook, but, I cannot get the answer to match that in the back of book and it's a simple problem.
Here it is, a cannon weighing 400 lb, fired a cannon ball weighing 10 lb, and had a muzzle velocity of 200 ft/s. For a 10 degree elevation angle, determine the velocity of the cannon after it was fired?
My immediate though was conservation of linear momentum, with Mc*Vcx + Mb+Vbx = 0, where the subscript c, b, and x are cannon, cannon ball, and horizontal direction, respectively.
Substituting the following values
Mc = 400lb
Mb = 10lb
Vbx = 200*sin(10^o)
and solving for Vcx
yields -4.9240 ft/s (sign is relative to coordinate system)
however the back of the book is stating -4.80 ft/s, WTF am I doing wrong?
Here it is, a cannon weighing 400 lb, fired a cannon ball weighing 10 lb, and had a muzzle velocity of 200 ft/s. For a 10 degree elevation angle, determine the velocity of the cannon after it was fired?
My immediate though was conservation of linear momentum, with Mc*Vcx + Mb+Vbx = 0, where the subscript c, b, and x are cannon, cannon ball, and horizontal direction, respectively.
Substituting the following values
Mc = 400lb
Mb = 10lb
Vbx = 200*sin(10^o)
and solving for Vcx
yields -4.9240 ft/s (sign is relative to coordinate system)
however the back of the book is stating -4.80 ft/s, WTF am I doing wrong?