Having trouble with relative motion

AI Thread Summary
To hit a battleship moving east at 24 km/h, a submarine must fire a torpedo traveling at 50 km/h at a specific angle. The torpedo needs to be aimed ahead of the battleship to account for its movement during the torpedo's travel time. The problem involves solving for the angle using relative motion equations and potentially applying calculus for related rates. A geometric approach can also be utilized to visualize the vectors involved. The correct firing angle for the torpedo is determined to be 46.8º east of north.
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Homework Statement



A battleship steams due east at 24 km/h. A submarine 4.0 km away fires a torpedo that has a speed of 50 km/h. If the ship is seen 20º east of north, in what direction should the torpedo be fired to hit the ship?


Homework Equations



x motion of ship: x = vt

x motion of torpedo: x' - x_{}0' = v_{}x't

y motion of torpedo: y - y_{}0 = v_{}yt

The Attempt at a Solution



The torpedo will hit the ship when its at y = 0, thus solve for t in the y motion equation of the torpedo.

0 - y_{}0 = v_{}yt

Then x = x'

In which you get x_{}0 + v_{}x (-y_{}0 / v_{}y) = x_{}0' + v_{}x' (-y_{}0 / v_{}y)

Do some algebra and you get x_{}0 - x_{}0' = (v_{}x' - v_{}x) / v_{}y

But I can't solve for \Theta
 
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The answer is 46.8º east of north...can someone help me how to handle this problem?
 
Hint: You must aim the torpedo ahead of the battleship in order for it to arrive where the battleship will be after the torpedo has made its run. You know the speeds of the battleship and the torpedo, as well as the relative bearing of the target from the sub at the moment of firing. You'll have to solve a related rate problem in order to obtain the complete firing solution.
 
Consider the battleship as the reference substance
 
azureth said:
Consider the battleship as the reference substance

So that means that the submarine is moving westward at 24 km/h and when the hypotenuse is 4 km, the torpedo is fired.

That means we use x^2 + y^2 = c^2.

d/dt (x^2 + y^2) = d/dt (c^2) which gives us 2x dx/dt + 2y dy/dt = 2c dc/dt...and since there is no change in y, the equation reduces to 2x dx/dt = 2c dc/dt.

x = 4 cos 70º
dx/dt = 24 km/h
I am not sure what c and dc/dt stand for...I am not sure what to solve. Sorry, I haven't done a related rate problem n a while.
 
I do not know how to solve this problem by using calculus.In fact you can draw the vectors and solve it geometrically.(I cannot post a picture for you now.If you still cannot work it out,I will reply to you a few days later)
 
SteamKing said:
Hint: You must aim the torpedo ahead of the battleship in order for it to arrive where the battleship will be after the torpedo has made its run. You know the speeds of the battleship and the torpedo, as well as the relative bearing of the target from the sub at the moment of firing. You'll have to solve a related rate problem in order to obtain the complete firing solution.

I tried the related-rate way a few posts above but I am not sure if I am doing it right.
 
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