Projectile motion submarine missle problem

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Homework Statement



A submarine-launched missile has a range of 4330 km.
(a) What launch speed is needed for this range when the launch angle is 45°?

(b) What is the total flight time?

(c) What would be the minimum launch speed at a 22° launch angle, used to "depress" the trajectory so as to foil a space-based antimissile defense?


Homework Equations



Y = 1/2gt2

And the other projectile motion equations.

The Attempt at a Solution



Well for part a: i convert the km to meters and find the time it would take for it to make it halfway:

2165000m = 1/2 (9.81m/s2)(t2)
t = 664.4s

Multiply that by 2 and get time of flight = 1328.7s

I get that far and it says it's already wrong because the time of flight is supposed to be 15.7 min. So I don't know what I am doing wrong. Could someone help me out with this problem?

Thanks
 

Answers and Replies

  • #2
NascentOxygen
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Have you seen a problem like this before? Are you allowing for the curvature of the earth?
 
  • #3
Orodruin
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You are equating the vertically travelled distance with half of the horizontal distance ...

Furthermore, 4000 km is a sizable distance in Earth terms. Approximating the gravitational field with a constant is just one of a large number of approximations that would have to be made to bring this problem to one of parabolic motion.
 
  • #4
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Curvature of the Earth does not need to be taken into account to obtain 15.7 min total flight time.

Going back to the relevant equations, what is "Y" and where does this equation come from?
 
  • #5
Orodruin
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Curvature of the Earth does not need to be taken into account to obtain 15.7 min total flight

I beg to differ. The height achieved with the middle point after 8 mins in a homogeneous gravitational field of 10 m/s^2 is of order 1000 km - a sixth of the Earth radius.
The initial vertical velocity required would be over 4 km/s - a third of Earth's escape velocity.

Furthermore, for the projectile to travel 4000 km, it is not coming back on a horizontal plane - it will have to go roughly an angle pi/5 around. That length scale alone tells you that curvature would matter.

And we have not even gotten started about air resistance at the velocities necessary to achieve this.
 
  • #6
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That length scale alone tells you that curvature would matter.

I am not debating this or any other points, which are well taken. Yet the expected answer assumes "flat Earth".
 
  • #7
Orodruin
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I am not debating this or any other points, which are well taken. Yet the expected answer assumes "flat Earth".


In that case I would like to have the problem maker as the lead robot and missile designer on the opposing side if my country should ever enter into a war... Posing questions and expecting answers based on approximations that are not in the very least bit viable seems anti-pedagogical to me. In particular when dressing the question up with statements of foiling space based defenses...
 
  • #8
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I don't think I have to worry about the curvature of the earth. This should be a basic projectile motion problem since thats what we are doing in our class.
 
  • #9
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So does any one know how i can go about the problem?
 
  • #10
NascentOxygen
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So it's not a missile, it's just a bullet! And the Earth isn't round, it's flat. Well, that simplifies things! :cool:

The range of a shell is its horizontal distance of travel, you have wrongly taken it to be vertical height.

The first step is to come up with an equation that relates launch speed to the horizontal distance of the projectile's flight. How will you do this?
 
  • #11
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The question he is referring to neither describes the Earth as "flat" nor "round" so it is hard to grasp these questions in the sense of the Physical world.

I am also working on this homework problem, and I've gotten question b) right, but not a). I also have not a clue what question c) is asking...
 
  • #12
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So it's not a missile, it's just a bullet! And the Earth isn't round, it's flat. Well, that simplifies things! :cool:

The range of a shell is its horizontal distance of travel, you have wrongly taken it to be vertical height.

The first step is to come up with an equation that relates launch speed to the horizontal distance of the projectile's flight. How will you do this?


X = v(cos 45)t

So i know x but i dont know v or time so if i have it right we have to do a system of equations. What should the second equation be?
 
  • #13
NascentOxygen
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X = v(cos 45)t

So i know x but i dont know v or time so if i have it right we have to do a system of equations. What should the second equation be?
That duration of horizontal flight is equal to the duration of the vertical flight. Exactly equal. :smile:

You are needing 2 equations, there being 2 unknowns.
 
  • #14
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R=sqrt((R*g*.001)/(sin(2*45)) is the formula for range.

Edit: the .001 is to convert the gravitational constant from m/s^2 to km/s^2.
 
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