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After blastoff, a space shuttle climbs vertically and a radartracking dish, located 1000 meters from the launch pad, follows the shuttle. How fast is the radar dish revolving 10 sec after blastoff if at that time the velocity of the shuttle is 100 m/sec and the shuttle is 500 meters above the ground?

h = height = 500 + 100 t
b = base = 1000
A = elevation angle. we want dA/dt
tan A = h/b = 100(5+t) / 1000 = .5 +.1 t
(1 /sec^2 A) dA/dt = 0 + .1
dA/dt = .1 sec^2 A
at this time ratio of b to h is 2/1
so hypotenuse is sqrt 5
cos A = 2/sqrt 5
sec A = sqrt 5/2
sec^2 A = 5/4
so
dA/dt = .1 (5.4) = .54 radians/second
or
31 degrees per second
sec A = 1/cos A = 1/(1000/(1000^2
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