A television camera is on the ground filming the lift-off of a space shuttle. The camera is 2000 feet from the

Question

A television camera is on the ground filming the lift-off of a space shuttle. The camera is 2000 feet from the launch pad. The shuttle is rising straight up at a rate given by the equation h= 50t^2 when h = height and t= time in seconds after liftoff. Find the rate of change in the angle of elevation from the camera to the shuttle .at exactly 10 seconds after lift off

Answer 1

Height h = 50t^2 = 5000 ft

Elev. angle theta = arctan(h/2000) = 1.1903 rad

Velocity v = derivative of 50t^2 = 100t = 1000 ft/s

Elev. angle rate of change = v/2000*cos(theta) = 0.1857 rad/s

Answer 2

make a triangle with on leg 2000 and one leg 5000 (50x10^2).  Then calculate tangent of 5000/2000