- #1

Stobbe

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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?