posted by :) on .
After blast-off, a space shuttle climbs vertically and a radar-tracking dish, located 1000 meters from the launch pad, follows the shuttle. How fast is the radar dish revolving 10 sec after blast-off 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
dA/dt = .1 (5.4) = .54 radians/second
31 degrees per second
sec A = 1/cos A = 1/(1000/(1000^2