At one moment during its flight a thrown basketball experiences a gravitational force of 1.5N down and air resistance of 0.40N(32 degree above horizontal). Calculate the magnitude and direction of of the net force of the ball

1. The magnitude of the net force is 1.33 N.

2. The direction is 14.8° with respect to the vertical.        

Explanation:

1. The magnitude of the net force is given by:

Where:

: is the sum of the forces acting in the x-direction

: is the sum of the forces acting in the y-direction

Let's find the forces acting in the x-direction and in the y-direction.

In the x-direction:

Where:

Fa: is the force of air resistance = -0.40 N. The negative sign is because this force is in the negative x-direction.  

θ: is the angle = 32°

In the y-direction:

Where:

: is the gravitational force = -1.5 N

Hence, the magnitude of the net force is:

2. The direction of the net force is:

The angle is 14.8° with respect to the vertical.

I hope it helps you!

your answer is, orbit around the nucleus

what is the actual question though?