Dropping the ball (HEEEEEEEELP)

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In summary, Dropping the ball! (HEEEEEEEELP!) requires knowing the final velocity, the acceleration due to gravity, and the mass of the object being dropped. To resist the deceleration, the person would need to apply a force of 293 pounds.
  • #1
sugarboyae
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Dropping the ball! (HEEEEEEEELP!)

Okay. It has been YEARS since I've done stuff like this but I need a rough number estimate for a project I'm doing. Here is the situation:

You have 20k wrecking ball you're dropping from 1m height. You're friend will catch it at the end of that 1m drop with only 0.15m to decelerate the mass to ZERO m/s.

What is the rate of deceleration??

AND... what is the equivalent force needed to resist that type of deceleration in that super small time?

Sorry if these questions aren't even valid. Thanks in advanced for the answers!
 
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  • #2
Final velocity V^2 = 2gs where g = 9.8m/s^2 s = distance = 1m

Deceleration is just, a = V/t
And Force F = m a
 
  • #3
^ so then.. what is the answer? this is a real life question haha...
 
  • #4
sugarboyae said:
this is a real life question haha...

All the more reason to learn the physics. :P
 
  • #5
<--- said:
All the more reason to learn the physics. :P

well that wasn't helpful at all.

actually I'm just years on end rusty... just need a bit of help if someone wouldn't mind? apparently everyone thinks its all easy here ... which is why this forum is here to help, yes?
 
  • #6
When you say 20k, do you mean 20,000 lbs, 20 kg, 20,000 kg??

I don't people are interested in giving you the answer. They are more interested in leading you to the answer. I'm happy to help.
 
  • #7
^ 20kg. sorry.

i've actually calculated out my own results previous to posting. but i'd like to have someone confirm it w/o bias.
 
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  • #8
First of all, I don't know why I asked about the mass because it's irrelevant. You have to do this in 2 steps. You will use the same equation 2 times. First, you need to find the final velocity (Vf) at the very bottom of the first meter. Then you can calculate the acceleration (a) by plugging in known values.

Vf^2 = Vi^2 + 2*a*s

Vf = Final Velocity
Vi = Initial Velocity
a = acceleration (due to gravity in this case, approx 9.8m/s^2)
s = distance (technically, it's the change in distance)

First, find the Vf (Remember Vi is 0):

Vf^2 = Vi^2 + 2*a*s
Vf = SQRT(2*9.8*1)

Second, you'll use the found value as you new Vi (Initial Velocity) and your Final Velocity is Zero.

Vf^2 = Vi^2 + 2*a*s
0 = SQRT(2*9.8*1)^2 + 2*a*.15

Then solve for a. Acceleration will be positive because it is accelerating (decelerating) upwards. This is hidden in the SQRT term.

As previously stated, you then use (a) and your mass (m) to find the force (F): F= m*a

If this isn't clear, please let me know. Also, what value did you get?

EDITED: OOPS! Wrote the equation wrong! I forgot to square the Vi and Vf terms. Also, be careful about your signs.
 
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  • #9
unless my parameters of xcel are busted, my decel was = 1283.148 m/s^2.

now the more complicated question is. how do we convert that into a force equivalent. because the final question is... if i drop that kind of weight and catch it myself. how much force am i truly resisting after letting that ball freefall for 1m and using only .15m of displacement for decel.
 
  • #10
subtract the initial velocity and then divide by 2*s

(2*9.8)/(2*.15) = a

Subtracting will make the term negative, but the acceleration due to gravity (9.8) is negative and cannot be used under the SQRT. We know it is positive because to slow it down, we have to accelerate the object upward.

a = 65.3 m/s^2

F = ma = 20 * 65.3 = 1306 N (Newtons)
N = kg * m/s^2


If you convert this force to pounds, you get:

F = ma = 1.37 * 214.2 = 293 lbs
lbs = sl * ft/s^2

EDIT: Sorry, I wasn't checking myself as I went. I made a few corrections.

:) I would move my hand before and let it hit the ground!
 
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  • #11
Well F =ma if that is the correct deceleration (I didn't check myself) that translates to 5774 pounds of force.
That doesn't sound very safe if you are planing on catching that in you arms.

Edit: Tineras posted the correct a/F, that's still 294 pounds of force...
 
  • #12
thanks everyone! yea ~ 300lbs was what my findings were as well. Thansk for all the help! Appreciate it greatly!
 

Related to Dropping the ball (HEEEEEEEELP)

1. What is the concept of "Dropping the ball"?

"Dropping the ball" is a metaphor for failing to follow through or complete a task or responsibility. It can also refer to making a mistake or error in judgment.

2. How does "Dropping the ball" affect scientific research?

In scientific research, "Dropping the ball" can have serious consequences as it can lead to inaccurate or unreliable results. It is important for scientists to be thorough and diligent in their work to avoid this.

3. What are some common reasons for "Dropping the ball" in a scientific experiment?

Some common reasons for "Dropping the ball" in a scientific experiment include lack of attention to detail, inadequate planning, and human error. It can also occur due to equipment failure or unexpected variables.

4. How can scientists prevent "Dropping the ball" in their research?

To prevent "Dropping the ball" in scientific research, scientists can take steps such as carefully planning their experiments, double-checking their work, and being open to feedback and collaboration with others. It is also important to constantly review and improve upon methods and procedures.

5. What are the consequences of "Dropping the ball" in the scientific community?

The consequences of "Dropping the ball" in the scientific community can range from minor setbacks to significant setbacks in advancing knowledge and understanding. It can also damage the credibility of the researcher and their work. Therefore, it is crucial for scientists to be diligent and accountable in their research.

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