Posted by jon on .
1. During World War 2, a U.S. navy submarine, sitting at a full stop on the surface of the Philippine Sea facing due north sights a Japanese Aircraft Carrier 5 miles, dead ahead. The carrier is sailing due east at a speed of 10 MPH. If the submarine carries torpedoes that run at 40 MPH, and it takes 3 minutes to aim, load, and fire the torpedo, at what angle must the submarine fire the torpedo to strike the carrier (report to 2 decimal places, i.e. 35.62 degrees). What is the running time of the torpedo? What is the total elapsed time from initial sighting to torpedo impact?
2.For the same data in problem number 1, show an alternate solution to the problem by finding the derivative of the trigonometric function, tan X for the given data. Show that the derivative of the tangent function will give the same angle found in the question 1. HINT: Angle measure must be done in radians.
Draw the picture.
The E distance component is 3/60*10 + 40Trunning. The N compoent is 5, the hypot is 40Trunning
Use the pyth theorm to get an equation, wit will be a quadratic , solve it for Trunning.
The N component is 5mi=40CosTheta*T, solve for Theta.
2. I don't understand this part. Theta is the firing angle, it is a constant, so the d/dt of tanTheta=0. Maybe the problem meant Theta as a function of something (initial range, carrier velocity, or torpedo velocity), but that is speculation. I am missing something here.
FYI, the best submarine torpedo in WWII was the 21inch Mk 17, it has a max speed of 33kts, and a max range of 13.5kyards. The torpedo cited above would have been a fantasy, as would have been a enemy carrier going only 10kts on a steady course. Life is not simple in war.