Calculate Impact Speed of Ball Bearings Dropped in Shot Tower

In summary: So v= -9.8 * 4.5= -44.1 m/s and as the ball falls from a height, v= u+at, so u= -44.1 m/s. In summary, ball bearings are made by dropping molten metal in a shot tower and letting it solidify as it falls. To solidify enough for impact, the bearing needs to fall from a height of 4.50 meters. The bearing's impact speed can be calculated using the kinematics equation v=u+at, with u=-44.1 m/s and t=4.5 seconds.
  • #1
Kalie
46
0
Ball bearings are made by letting spherical drops of molten metal fall inside a tall tower called a shot tower and solidify as they fall.
If a bearing needs 4.50 to solidify enough for impact, how high must the tower be?
What is the bearing's impact speed?

I got the answer for the first part correct it was 99.2
d=.5*9.8*4.5^2=99.2
but I am not sure how to approach the second part...

What is the bearing's impact speed?
I search all over but couldn't find an equation could someone help me with that.
All I need is the equation...thank you
 
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  • #2
Look for the kinematics equations. Given all the information you have, you actually have a choice on the equation you want to use.

How about the intuitive: final velocity = initial velocity + (acceleration x time)
 
  • #3
Kalie said:
Ball bearings are made by letting spherical drops of molten metal fall inside a tall tower called a shot tower and solidify as they fall.
If a bearing needs 4.50 to solidify enough for impact, how high must the tower be?
What is the bearing's impact speed?

I got the answer for the first part correct it was 99.2
d=.5*9.8*4.5^2=99.2
but I am not sure how to approach the second part...

What is the bearing's impact speed?
I search all over but couldn't find an equation could someone help me with that.
All I need is the equation...thank you

You know that v= -9.8 t m/s2 and t= 4.5 s!
 

1. How do you calculate the impact speed of ball bearings dropped in a shot tower?

The impact speed of ball bearings dropped in a shot tower can be calculated using the formula: v = √(2gh), where v is the impact speed, g is the acceleration due to gravity, and h is the height from which the ball bearings are dropped.

2. What is the acceleration due to gravity?

The acceleration due to gravity is the rate at which objects fall towards the Earth's surface. It is typically denoted by the symbol g and has an approximate value of 9.8 m/s² on Earth.

3. How does the height of the shot tower affect the impact speed of the ball bearings?

The impact speed of the ball bearings increases as the height of the shot tower increases. This is because the ball bearings have a longer distance to fall, thus increasing their velocity due to the acceleration of gravity.

4. What other factors can affect the impact speed of the ball bearings?

The impact speed of the ball bearings can also be affected by air resistance, the shape and size of the ball bearings, and the surface on which they land. These factors can cause variations in the calculated impact speed.

5. Are there any safety precautions that should be taken when conducting this experiment?

Yes, it is important to wear appropriate safety gear, such as goggles and gloves, when dropping ball bearings in a shot tower. It is also important to conduct the experiment in a controlled environment to prevent any accidents or injuries. Additionally, make sure to follow proper handling and disposal procedures for the ball bearings after the experiment is complete.

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