Missile Defense System You are designing a missile defense system that will shoo

Question

Missile Defense System 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, of the incoming missile. Your Patriot Rocket is set to fire at an angle of = 65.0 degrees from vertical. You design the Patriot Rocket so the magnitude of its acceleration is given by: where A can be set on your Patriot Rocket as it is fired, b-0.35 s-1, and t-0 when it is fired. The direction of your Patriot Rocket's vector acceleration stays at the same angle, , for the entire trip. If an incoming missile passes over the defense base at a height of 5.40 km and at a constant speed of 741.0 m/s (this means that v, is constant), solve for the value of A 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 horizontal distance Ar from the launch station to the impact position, and the time , at impact. a, = 1.158 x 104 m 1pts You are correct. Your receipt no. is 150-4756 -1.563x 101 s 1pts You are correct. our receipt no. is 150-3221Previous Tries -3716.06 m/s2 Submit Answer Incorrect. Tries 9/15 Previous Tries 1pts

Explanation / Answer

We have to use the acceleration function to find the velocity function of the missile :

         v = a dt =    Ce-bt dt = - (C/b) e-bt + K     where K is an arbitrary constant.

We can find K by using the fact that when t = 0, the velocity is zero. So K = C/b    and

          v = ( 1 -e-bt)         where   = C/b

Now, we recognize that velocity and acceleration are in thesame direction. And the horizontal velocity is simply the magnitudeof the total velocity times cos. So,

horizontal velocity = vx =   cos (1 -e-bt)   

We also know that the horizontal displacement of the Patriot missile is....

        x =   vx dt  =    cos (1 -e-bt) dt  = cos (1 -e-bt) = cos (t + (1/b)e-bt ) = ( cos / b) ( bt +e-bt )   

If we eval this from t = 0 to T where T is the time of impact, we get:

x = (C cos / b2 ) ( b T + e-bT   -   1 )

Notice in this expression there is only one unknown... the value of C. You have values for everything else. So we can nowplug in and get C:

1.158 x 104 = (C cos65° /0.352 ) (0.35 * 15.63 + e-0.35*15.63 - 1)

C = 750.12 m/s2

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