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olga_2 [115]
3 years ago
14

Amy throws a softball through the air. What are the different forces acting on the ball while it’s in the air? The softball expe

riences force as a result of Amy’s throw. As the ball moves, it experiences from the air it passes through. It also experiences a downward pull because of .
Physics
2 answers:
sasho [114]3 years ago
7 0
<h2>The different forces acting on the ball while its in air</h2>

Amy throws a softball through the air. Applied, drag and gravitational forces are acting on the ball while it’s in the air. The softball experiences force as a result of Amy’s throw. As the ball moves, it experiences from the air it passes through.

It also experiences a downward pull because earth has the property to attract everything which is on the earth towards it. The ball is moving in the air but earth applies force on the ball to get back on the ground. Hence, in this way, gravitational force applies.

There is also a drag force which results due to friction that is present in the air. It resist to move ball in the air and there will also be applied force which is given by a person who throws by applying force.    

ivolga24 [154]3 years ago
7 0

Answer:

The downward pull is the gravitational pull of the earth.

Explanation:

The ball experiences three types of forces.

1. Applied force by Amy.

2. Dragging force due to the air friction which resists the motion of the ball.

3. The gravitational pull which acts in downward direction applied by the earth.

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Answer:

F₃ = 122.88 N

θ₃ = 20.63°

Explanation:

First we find the components of F₁:

For x-component:

F₁ₓ = F₁ Cos θ₁

F₁ₓ = (50 N) Cos 60°

F₁ₓ = 25 N

For y-component:

F₁y = F₁ Sin θ₁

F₁y = (50 N) Sin 60°

F₁y = 43.3 N

Now, for F₂. As, F₂ acts along x-axis. Therefore, its y-component will be zero and its x-xomponent will be equal to the magnitude of force itself:

F₂ₓ = F₂ = 90 N

F₂y = 0 N

Now, for the resultant force on ball to be zero, the sum of x-components of the forces and the sum of the y-component of the forces must also be equal to zero:

F₁ₓ + F₂ₓ + F₃ₓ = 0 N

25 N + 90 N + F₃ₓ = 0 N

F₃ₓ = - 115 N

for y-components:

F₁y + F₂y + F₃y = 0 N

43.3 N + 0 N + F₃y = 0 N

F₃y = - 43.3 N

Now, the magnitude of F₃ can be found as:

F₃ = √F₃ₓ² + F₃y²

F₃ = √[(- 115 N)² + (- 43.3 N)²]

<u>F₃ = 122.88 N</u>

and the direction is given as:

θ₃ = tan⁻¹(F₃y/F₃ₓ) = tan⁻¹(-43.3 N/-115 N)

<u>θ₃ = 20.63°</u>

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