Suppose three forces are acting on a model rocket launching into the air. The force of gravity is 1.2 N acting downward. The rocket engine has a force of 12.3 N acting at an angle of 65⁰ above the horizontal to the right. The wind pushes the rocket with a force of 2.5 N to the left. Find the magnitude and direction of the resultant force on the rocket. Please answer!!! Offering 50 points for answer and explanation

Decompose the forces acting on the rocket in horizontal and vertical components.

• net horizontal force:

∑ F = (12.3 N) cos(65°) - 2.5 N ≈ 2.7 N

(notice we're taking "to the right" to be the positive direction)

• net vertical force:

∑ F = (12.3 N) sin(65°) - 1.2 N ≈ 9.9 N

The resultant force then has magnitude

√((2.7 N)² + (9.9 N)²) ≈ 10.3 N

Since the horizontal and vertical components of the resultant are both positive, it points at an angle between 0° and 90° from the positive horizontal, so that

tan(θ) ≈ (9.9 N) / (2.7 N) ≈ 3.69

===>   θ ≈ arctan(3.69) ≈ 75°

1 a

2 b

3 b

4 d

5 d

6 d

explanation:

it is very well related to a lot of things but with my limited knowledge i can confidently say that the answer is space.