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Direction and magnitude of the average acceleration

  • Thread starter greydog
  • Start date
5
0
1. A ball is dropeed from a height of 2.5m and rebounds to a height of 2.1m. If the ball is in contact with the floor for .70ms determine 1: direction and 2: the magnitude of the balls average acceleration due to the floor

I know the equation for average acceleration = change in velocity/ time to make change

I found initial velocity by: v(initial)= sqrt(2*9.8*2.5) = 7m/s
I found final velocity by: v(final)= sqrt(2*9.8*2.1) = 6.42m/s

When i plug those into the avg acceleration equation i have: (6.42m/s - 7 m/s)/.0007s
to get: -828.57m/s So the direction would be down?

Am i doing my math right I am not sure where the negatives go for gravity either. Thanks for your help.
 
133
0
1. A ball is dropeed from a height of 2.5m and rebounds to a height of 2.1m. If the ball is in contact with the floor for .70ms determine 1: direction and 2: the magnitude of the balls average acceleration due to the floor

I know the equation for average acceleration = change in velocity/ time to make change

I found initial velocity by: v(initial)= sqrt(2*9.8*2.5) = 7m/s
I found final velocity by: v(final)= sqrt(2*9.8*2.1) = 6.42m/s


Okay, good start here. But you need to pick a sign convention. Usually we assume that positive is up. See what difference that makes.
 
5
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do you mean initial velocity should be: v(initial) = sqrt(2*-9.8m/s^2*-2.5m) = 7m/s
and then final velocity would be the same, if so then am i right about the direction being down because the average acceleration is - ? Thanks for helping.
 

lewando

Homework Helper
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Unfortunately, when you take the square root of a number you get a positive and negative result. This means you have to think about which of the two results applies to your system. Using Ignea_unda's recommendation, what would be the sign of the magnitude of the velocity upon initial contact with the ground?

What would be the sign for the velocity as the ball stops contacting the ground?
 
5
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Upon initial contact with the ground the sign should be positive, i would think. When the ball hits the ground it is 0. As the ball moves upward, gravity is working on it again, but the ball is moving positively. What am I doing wrong here? It's just not clicking. I hope i'm close though thanks.
 

lewando

Homework Helper
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Upon initial contact with the ground the sign should be positive, i would think.
The ball is moving downward-- it has downward velocity.

When the ball hits the ground it is 0.
This is true when the ball "bottoms out", somewhere in the .70ms interval.

As the ball moves upward, gravity is working on it again, but the ball is moving positively.
Yes, it is moving in the positive, up, direction. The velocity is pointing up.

You are close. Just get the signs of the 2 velocities right.
 
5
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I found this on your forum:

If velocity is (-) (down or to the left) and x (motion) is speeding up, then acceleration will have a (-) sign in front of it.
If velocity is (+) (up or to the right) and x (motion) is slowing down, then acceleration will have a (-) sign in front of it.

So my v(initial) should have negative sign in front and my v(final) should have a positive sign. correct? Then my avg acceleration = 6.42m/s-(-7m/s)/.0007 s giving a positive acceleration therefore a positive direction yes? I hope.
 

lewando

Homework Helper
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133
Yep, good work by you!
 
5
0
Yay! Thank you so much for your help! Have a great day!
 

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