Question

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

Answer 1

Answer:

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:

$|F| = \sqrt{F_{x}^{2} + F_{y}^{2}}$

Where:

$F_{x}$: is the sum of the forces acting in the x-direction

$F_{y}$: 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:

$\Sigma F_{x} = -F_{a}cos(\theta)$

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°

$\Sigma F_{x} = -0.40 N*cos(32) = -0.34 N$

In the y-direction:

$\Sigma F_{y} = F_{g} + F_{a}sin(\theta)$

Where:

$F_{g}$: is the gravitational force = -1.5 N

$\Sigma F_{y} = -1.5 N + 0.40 N*sin(32) = -1.29 N$

Hence, the magnitude of the net force is:

$|F| = \sqrt{(-0.34 N)^{2} + (-1.29)^{2}} = 1.33 N$

2. The direction of the net force is:

$tan(\alpha) = \frac{F_{x}}{F_{y}} = \frac{-0.34}{-1.29}$

$\alpha = tan^{-1}(\frac{-0.34}{-1.29}) = 14.8$

The angle is 14.8° with respect to the vertical.

I hope it helps you!