What is the speed when a ball hits the ground

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SUMMARY

The discussion focuses on calculating the speed of a sphere with mass m and radius R when it hits the ground, factoring in quadratic air resistance while neglecting linear air resistance. The relevant equation derived is mg - cv² = m(dv/dt), leading to the final velocity expressed as v(t) = v_terminal * tanh(gt/v_terminal). Specific parameters include R = 0.2c, v0 = 1.3 m/s, and m = 100 mg. The analysis emphasizes the necessity of considering both upward and downward motion to accurately determine the speed just before impact.

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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\frac{dv}{dt}


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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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.
 
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.
 
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?
 

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