# Rocket Force - Max Range vs. Time

1. Sep 20, 2007

### bphysics

Rocket Force -- Max Range vs. Time

1. The problem statement, all variables and given/known data

"Assume an ICBM of maximum range 8000 km is fired at your city from a distance of 8000 km."

1) How much warning time will you have if the missile is first detected at its halfway point
2) How fast will the missile be travelling when it is first detected?
3) With what velocity will it strike its target?
4) Now assume an ICBM of greater maximum range is fired at your city from a distance of 8000 km at an angle of 50 degrees. How much more warning time will you have if the missile is detected at the halway point than the first scanario?

2. Relevant equations

x = Vx0t
y = Vy0t - (1/2)gt^2

v = v0 + at
v^2 = v(2/0) + 2a(x - x0)
x = x0 + v0t + (1/2)at^2

3. The attempt at a solution

I keep on trying to figure out how to even start off this sucker. I've performed projectile motion problems before, but this "appears" more complex -- in the sense that I have no initial velocity and no initial angle. To me, I am uncertain how to proceed with this problem.

It would appear that I need to utilize the maximum range of 8000 km combined with my knowledge that gravity is taking this missle down at 9.8 m/s^2 to help me calculate for these other values, but I don't seem to grasp how to.

I'll be writing more of this on my whiteboard, and if I discover anything, I'll post it.

For now, I'm just hoping someone can put me on the right track.

2. Sep 20, 2007

### FedEx

Apply the equation for the range of the rocket. From here you will get the initial velocity.

3. Sep 21, 2007

### bphysics

FedEx, I don't seem to be able to utilize any of the equations which I have listed at this time due to how I do not have enough information to "complete" the variable requirements enough to solve for a variable.

Can you tell me how you can see this being solved, and more, specifically, via which listed equation you think I have enough evidence for? I may simply be looking at the problem from the wrong angle.

4. Sep 21, 2007

### learningphysics

Try to derive a formula for the range, in terms of v0 and theta. Hint: what is the time to reach the maximum height.... if you double that... you get the time it hits the range.

5. Sep 21, 2007

### bphysics

Understood - however, your discusing time here.

I have no time data.. unless I am just very confused right now.

6. Sep 21, 2007

### FedEx

I thought that as you have solved a few projectile problems you would be knowing the equations of projectile motion.

No matter. keep the following in mind.

1) The horizontal range of the projectile is max when angle is 45

2) The horizontal range is given by R = $$\frac{v^2\sin2\theta}{g}$$

3) The time taken by the projectile to reach the max height is equal to the time taken by it to come to the ground from the max height.

4) The max height is given by $$H_{max}$$ = $$\frac{v^2sin^2\theta}{g}$$

5)The total time is given by T = $$\frac{2vsin\theta}{g}$$

Here v is the initial velocity.