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After blast off, a space shuttle climbs vertically and a radar tracking dish, located 1000 yd from the launch pad, follows the shuttle. How fast is the radar dish revolving 10 sec after blast-off if at the time the velocity of the shuttle is 100 yd/sec and the shuttle is 500 yd above the ground?

 Dec 14, 2014

Best Answer 

 #1
avatar+118703 
+10

This is my best guess

 

When t=10sec  y=500yards   and dy/dt=100 yards/sec   and    θ=atan(5001000)=atan(0.5)

 

tanθ=y1000y=1000tanθdydθ=1000sec2θdθdy=cos2θ1000dθdt=dθdy×dydtWhent=10dθdt=cos2θ1000×100dθdt=cos2(atan(0.5))10dθdt=45÷10dθdt=450dθdt=225radians/sec 

 

 

 

I think that this is correct.  

 Dec 14, 2014
 #1
avatar+118703 
+10
Best Answer

This is my best guess

 

When t=10sec  y=500yards   and dy/dt=100 yards/sec   and    θ=atan(5001000)=atan(0.5)

 

tanθ=y1000y=1000tanθdydθ=1000sec2θdθdy=cos2θ1000dθdt=dθdy×dydtWhent=10dθdt=cos2θ1000×100dθdt=cos2(atan(0.5))10dθdt=45÷10dθdt=450dθdt=225radians/sec 

 

 

 

I think that this is correct.  

Melody Dec 14, 2014
 #2
avatar+130477 
+5

Yep....nice job !!!

 

 Dec 14, 2014
 #3
avatar+118703 
0

Thanks Chris :)

 Dec 14, 2014

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