What is the speed when a ball hits the ground

In summary, the conversation discusses finding the speed of a sphere when it hits the ground after being fired straight up with a given mass, radius, and initial velocity. The question also addresses the inclusion of air resistance and the need to break the motion into two parts. The conversation also mentions the concept of terminal velocity and the importance of knowing the starting height to determine the final speed.
  • #1
chris_0101
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0

Homework Statement



A sphere of a mass m and radius R is fired straight up with a speed v0
a) What is its speed when it hits the ground (Include quadratic air resistance but ignore linear air resistance

b) Evaluate the result for part (a) with R = 0.2c, v0 = 1.3m/s and m = 100mg

I am puzzled with regards to whether or not I should include the upward motion and solving the question entirely

Homework Equations



mg -cv2 = m[itex]\frac{dv}{dt}[/itex]


The Attempt at a Solution



After following my class notes, I found that the final velocity as a function of time is:
v(t) = vterminal*tanh(gt/vterminal})

Any help is greatly appreciated,

thanks
 
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  • #2
I am puzzled with regards to whether or not I should include the upward motion and solving the question entirely
How could you determine the extent to which the upwards part of the motion is important? What would this determine? Do you need it?

Note: air resistance will increase acceleration when going up and decrease acceleration going down, so whatever you do you'll want to break the motion into two parts.
 
  • #3
Wait, how can air resistance increase acceleration when going up. From what I have understood, air resistance slows down the object flying upward, likewise with gravity.

I do understand that I have to break up the question into the upward and downward parts, but the question asks for the speed just before it hits the ground. Furthermore, the object being shot with an initial velocity will eventually go to rest at its highest point, then fall back down again. This is what caused me to believe that the upward component is negligible.
 
  • #4
acceleration is a vector.

Without air resistance, the only force is gravity (weight), pointing down.
The acceleration vector, therefore, points downwards.

When the ball goes up, the force of the air resistance points down, adding to the force of gravity and so increasing the acceleration.

When the ball goes down, the air resistance points up, subtracting from the force of gravity and so reducing the acceleration.When the ball falls from it's highest point to the ground, what determines the speed when it strikes?
Perhaps the ball has enough time to reach it's terminal velocity? How would you know?
Don't you need to know how high it started from?
 
  • #5


I would like to clarify that the speed of a ball when it hits the ground depends on several factors such as the initial velocity, air resistance, and the height from which it was dropped. In this case, we are given the initial velocity and mass of the ball, as well as the radius of the ball. However, we are not given the height from which it was fired, so it is difficult to accurately determine the speed when it hits the ground.

In order to solve this problem, we need to use the equations of motion and include the effects of air resistance. The equation you have used, v(t) = vterminal*tanh(gt/vterminal}), is the equation for the terminal velocity, which is the maximum speed that an object can reach when falling due to the balance of gravity and air resistance. However, this equation does not take into account the initial velocity of the ball.

To solve for the final velocity when the ball hits the ground, we can use the equation:

v(t) = v0 - gt - (c/m)v2t

Where v0 is the initial velocity, g is the acceleration due to gravity, c is the coefficient of air resistance, and m is the mass of the ball. We can rearrange this equation to solve for t, the time it takes for the ball to hit the ground:

t = (v0 - v(t))/g - (c/m)v2t

Once we have t, we can plug it into the original equation to solve for the final velocity:

v(t) = v0 - gt - (c/m)v2t

In the case of part (a), where we are ignoring linear air resistance, the equation becomes:

v(t) = v0 - gt

To solve for the final velocity, we need to know the initial velocity, which is v0 in this case. However, in part (b), we are given the radius of the ball, which is 0.2m, and the mass of the ball, which is 100mg. In order to solve for the initial velocity, we need to know the height from which the ball was fired.

In conclusion, to accurately determine the speed when a ball hits the ground, we need to know the initial velocity, the height from which it was fired, and the effects of air resistance. Without this information, it is difficult to provide a specific answer to this question.
 

FAQ: What is the speed when a ball hits the ground

What is the speed when a ball hits the ground?

The speed of a ball when it hits the ground depends on several factors such as the initial height from which it was dropped, the gravitational acceleration, and the air resistance. The speed can be calculated using the equation v = √(2gh), where v is the speed, g is the gravitational acceleration (9.8 m/s^2), and h is the initial height.

Does the speed of a ball when it hits the ground change if it is dropped from different heights?

Yes, the speed of a ball when it hits the ground will change if it is dropped from different heights. This is because the initial height affects the potential energy of the ball, which is then converted into kinetic energy as it falls. Therefore, the higher the initial height, the greater the potential energy and the faster the ball will be when it hits the ground.

How does air resistance affect the speed of a ball when it hits the ground?

Air resistance plays a significant role in determining the speed of a ball when it hits the ground. As the ball falls, it experiences air resistance, which opposes its motion and causes it to slow down. This means that the speed of the ball when it hits the ground will be lower than the calculated speed due to air resistance.

Is the speed of a ball when it hits the ground affected by the mass of the ball?

Yes, the speed of a ball when it hits the ground is affected by its mass. According to Newton's Second Law of Motion, the acceleration of an object is directly proportional to the force applied and inversely proportional to its mass. Therefore, a heavier ball will experience a greater force due to gravity and will fall faster, resulting in a higher speed when it hits the ground.

Can the speed of a ball when it hits the ground be greater than the speed at which it was dropped?

No, the speed of a ball when it hits the ground cannot be greater than the speed at which it was dropped. This is because of the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred from one form to another. Therefore, the total energy of the ball at any point during its fall cannot exceed the initial energy it had when it was dropped.

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