When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how fast is the angle of elevation of the rocket increasing, as seen by an observer on the ground 5 kilometers from the launching pad

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer (), measured in kilometers, can be represented by a right triangle:

(1)

Where:

- Horizontal distance between the rocket and the observer, measured in kilometers.

- Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket (), measured in sexagesimal degrees, is defined by the following trigonometric relation:

(2)

If we know that , then the expression is:

And the rate of change of this angle is determined by derivatives:

Where:

- Rate of change of the angle of elevation, measured in sexagesimal degrees.

- Vertical speed of the rocket, measured in kilometers per hour.

If we know that and , then the rate of change of the angle of elevation is:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5

0.11 m/s and 0.28 m/s should be correct