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.

The magnitude of the linear velocity of the rocket at the given position is 1,667.2 ft/s.

The magnitude of the acceleration of the rocket at the given position is 90.73 ft/s².

The given parameters;

The linear velocity and final angular velocity of the rocket is calculated as follows;

The radial acceleration of the rocket is calculated as follows;

The tangential acceleration is calculated as follows;

The resultant acceleration of the rocket is calculated as follows;

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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

c. newton's third law of motion