The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give the values r = 32700 ft, r¨ = 85 ft/sec2, and θ˙ = 0.019 rad/sec. Calculate the magnitudes of the velocity and acceleration of the rocket at this position.

velocity = 1527.52 ft/s

Acceleration = 80.13 ft/s²

Explanation:

We are given;

Radius of rotation; r = 32,700 ft

Radial acceleration; a_r = r¨ = 85 ft/s²

Angular velocity; ω = θ˙˙ = 0.019 rad/s

Also, angle θ reaches 66°

So, velocity of the rocket for the given position will be;

v = rθ˙˙/cos θ

so, v = 32700 * 0.019 / cos 66

v = 1527.52 ft/s

Acceleration is given by the formula;

a = a_r/sinθ

For the given position,

a_r = r¨ - r (θ˙˙) ²

Thus,

a = (r¨ - r (θ˙˙) ²) / sinθ

Plugging in the relevant values, we obtain;

a = (85 - 32700 (0.019) ²) / sin66

a = (85 - 11.8047) / 0.9135

a = 80.13 ft/s²