Calculating Final velocity and distance

[Burke]

Homework Statement

A ball, thrown upward, leaves the thrower's hand at a height of 0.8m with a velocity of 5m/s.
How fast is the ball moving just before it hits the ground? (assume the thrower moves out of the way)?

Homework Equations

Could someone check my work to make sure I have the correct answer to the problem? Thanks very much in advance. Note that my answer is a negative number. That doesn't seem right to me.

The Attempt at a Solution

How fast is the ball moving just before it hits the ground?[/B]
v = u + at
I calculated time as being 1.16
v= 5 + (-9.8)(1.16) = - 6.37
final velocity is - 6.37

 

[jbriggs444]

(Final Velocity)^2 = (Initial Velocity) ^2 + 2(Acceleration)(Displacement)
vf^2=5*2(-9.8)(0.8)= 8.85

How did you get from the first equation to the second? There is a plus sign missing. Also, consider the sign on your figure for displacement.

Edit: You also played fast and loose by taking the square root without telling anyone. The notation a=b=c means that all three terms are equal. Writing $v_f^2$ = some formula = $\sqrt{that\ formula}$ is incorrect.

 

[andrewkirk]

(Final Velocity)^2 = (Initial Velocity) ^2 + 2(Acceleration)(Displacement)
vf^2=5*2(-9.8)(0.8)= 8.85

Something has gone wrong here. In the first line there are two terms on the right-hand side, but then in the second line there is only one. It looks like you are multiplying the two terms rather than adding them.

 

[Burke]

How did you get from the first equation to the second? There is a plus sign missing. Also, consider the sign on your figure for displacement.

Edit: You also played fast and loose by taking the square root without telling anyone. The notation a=b=c means that all three terms are equal. Writing $v_f^2$ = some formula = $\sqrt{that\ formula}$ is incorrect.

Thanks for the input. I went back to the drawing board and redid the problem using a simpler formula a friend shared. Could you take another look at it? Thanks so much!

 

[Burke]

Something has gone wrong here. In the first line there are two terms on the right-hand side, but then in the second line there is only one. It looks like you are multiplying the two terms rather than adding them.

Thanks for the input. I went back to the drawing board and redid the problem using a simpler formula a friend shared. Could you take another look at it? Thanks so much!

 

[jbriggs444]

I calculated time as being 1.16

Rather than pulling 1.16 seconds out of thin air, you should justify that figure.

Then you should go back and see if the formula $v_f^2 = u^2 + 2as$ gives the same result for $v_f$

 

[Burke]

Thanks so much for not only the

Rather than pulling 1.16 seconds out of thin air, you should justify that figure.

Then you should go back and see if the formula $v_f^2 = u^2 + 2as$ gives the same result for $v_f$

Thanks so much for not only the help but for helping me to understand. In the formula v^2f = u^2+2as you provided me, would s=.8 in this particular situation? Determining the appropriate numbers to plug into the formulas is one of the areas I get tangled up,

 

[jbriggs444]

In the formula v^2f = u^2+2as you provided me, would s=.8 in this particular situation? Determining the appropriate numbers to plug into the formulas is one of the areas I get tangled up,

The sign convention is important. The displacement is downward. Your acceleration (a=-9.8) means that you have chosen down to be negative. Accordingly, the displacement is -0.8.

Mnemonic: The acceleration is in the same direction as the displacement. It will increase velocity.

 

[Burke]

Ahhhh..knowing acceleration and displacement should always be in the same direction helps. I recalculated using the formula you gave me and came up with almost the same answer (with the exception of a sign change). I know I am being a pain in the #%# but can you take one more look at this? This is one of those quizzes we can take a certain number of times and this is the only problem out of 30 problems that I am stuck on. I could just get the answer from a friend but I know that won't help me understand this for a test.

v^2f = u^2+2as

v^2f = 5^2 + 2(-9.8)(-.8) v^2 = 25 + 15.68 v^2=40.68 v= 6.378
So would answer be ball is moving at 6.378 m/s just before hitting ground?

 

[jbriggs444]

v^2f = 5^2 + 2(-9.8)(-.8) v^2 = 25 + 15.68 v^2=40.68 v= 6.378
So would answer be ball is moving at 6.378 m/s just before hitting ground?

Note that this matches the answer that you [currently] show in the original post. But that answer is negative and this answer is positive.

What could be up with that? Note that you determined that v^2 = 40.68. That is consistent with either v=6.378 or v=-6.378. The equation tells you the speed. It is up to you to figure out whether it is downward (negative velocity) or positive (positive velocity).

 

[Burke]

Ok I understand why v^2 = 40.68 can either be a positive or negative number. Given that the question doesn't state the downward is a positive direction, and I have chosen acceleration and displacement to be negative, my (limited) educated guess is that the velocity answer should be negative as well. So would the answer be -6.378 m/s?

 

[jbriggs444]

Ok I understand why v^2 = 40.68 can either be a positive or negative number. Given that the question doesn't state the downward is a positive direction, and I have chosen acceleration and displacement to be negative, my (limited) educated guess is that the velocity answer should be negative as well. So would the answer be -6.378 m/s?

I agree.

Either answer could be correct depending on the sign convention that is chosen. The problem statement gives a clue to the intended convention. "A ball, thrown upward ... velocity 5 m/s^2". If +5 m/s^2 is upward then a downward final velocity would be expressed as a negative number.

 

[Burke]

Thanks so much. You have been so helpful. I would love to give some positive feedback for you if you can tell me how.

 

[jbriggs444]

Thanks so much. You have been so helpful. I would love to give some positive feedback for you if you can tell me how.

You've already provided it. The "Like" pushbutton works too.

 

[Burke]

done... thanks so much