Question

In: Physics

You are designing a missile defense system that will shoot down incoming missiles that pass over...

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, hh, and speed, v1, of the incoming missile. Your Patriot Rocket is set to fire at an angle of θ=58 from vertical. You design the Patriot Rocket so the magnitude of its acceleration is given by:

a2=Ae-bt

where A can be set on your Patriot Rocket as it is fired, b=0.35 s-1, and t=0 s when it is fired. 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 4.8 km and at a constant speed of 682 m/s (this means that v⃗1 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 Δx from the launch station to the impact position, and the time tf at impact.

Find the horizontal distance to the collision

Find the time of collision

Find the value of rocket constant A

Solutions

Expert Solution

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 = β ( 1 -e-bt) where β = 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 cosθ. So,

horizontal velocity = vx = β cosθ (1 -e-bt)   

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

Δx = ∫vx dt = ∫β cosθ (1 -e-bt) dt = β cosθ ∫ (1 -e-bt) = β cosθ (t + (1/b)e-bt ) = (β cosθ / b) ( bt +e-bt )   

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

Δx = (C cosθ / 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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