Physics 2D kinematics problem, w/rockets

  • Thread starter ebbybeh
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  • #1
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Homework Statement


You are designing a missile defense system that will shoot down incoming missiles that pass over a perimeter defense post. The set-up is shown below. An incoming missile passes directly above the defense base. Radar at the base can measure the height, h, and speed, v1, of the incoming missile. Your Patriot Rocket is set to fire at an angle of Theta= 53.0 degrees from vertical. You design the Patriot Rocket so the magnitude of its acceleration is given by:
a=Ce^−bt
where C can be set on your Patriot Rocket as it is fired, and b = 0.40 s^-1. The direction of your Patriot Rocket's vector acceleration stays at the same angle, θ, for the entire trip. If an incoming missile passes over the defense base at a height of 4.80 km and at a constant speed of 795.0 m/s (this means that v1 is constant), solve for the value of C your Patriot Rocket must have in order to hit the incoming missile. You will also need to enter results from intermediate steps of your calculation, including the time t in between launch and impact, and the horizontal distance x from the launch station to the impact position.

Homework Equations


a=Ce^−bt

-The usual kinematics equation


The Attempt at a Solution


I tried this in a couple different ways for the past hour with no success. The first way was to use the position kinematic equation for the y direction, but since the acceleration formula has two variables i couldn't go further. I tried it from the incoming rockets y position but since we assume its staying in a straight line, i got t=0. I don't know how to integrate the acceleration formula above since there are two variables. Any tips on solving would be much appreciated.
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi ebbybeh! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)

With motion at an angle to gravity, just use the usual constant acceleration equations twice, once horizontally (with a = 0, of course) and once vertically (with a = g). :smile:
 
  • #3
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But doesn't the equation provide you with acceleration, or do you not have to use it
 
  • #4
tiny-tim
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(just got up :zzz: …)
But doesn't the equation provide you with acceleration, or do you not have to use it
oh, sorry :redface:, I should have said: add g to the vertical component of the rocket's acceleration, but use the horizontal component on its own …

the important thing is to use the usual constant acceleration equations twice. :smile:

(actually, looking at the question again, I'm not even sure that's right … is Ce−bt just the acceleration from the engines, so that you have to add g to it, vertically, or is it the total acceleration, including gravity? :confused:)
 
  • #5
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(just got up :zzz: …)


oh, sorry :redface:, I should have said: add g to the vertical component of the rocket's acceleration, but use the horizontal component on its own …

the important thing is to use the usual constant acceleration equations twice. :smile:

(actually, looking at the question again, I'm not even sure that's right … is Ce−bt just the acceleration from the engines, so that you have to add g to it, vertically, or is it the total acceleration, including gravity? :confused:)
I'm not sure either, but I was completely replacing the acceleration in both the x and y directions for Ce^-bt. I'll try again and see what happens:)
 

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