Homework Help: How fast was the ball thrown?

1. Nov 18, 2013

Medgirl314

1. The problem statement, all variables and given/known data

A boy throws a baseball directly upward. He catches it 3.2 seconds later.

2. Relevant equations

a=Δv/Δt

3. The attempt at a solution

Letting up be positive, I think the acceleration would be positive 9.8 m/s^2. I think I am trying to find the final velocity. I think that the inital velocity would be 0, right before he throws the ball. Could I go about this problem by using a=Δv/Δt to find velocity? This method seems to yield an answer of 3.0625 m/s. Thanks in advance for a helpful reply!

Last edited: Nov 18, 2013
2. Nov 18, 2013

Staff: Mentor

If up is positive, gravity's acceleration is downward, so it is negative.

What is the equation for the vertical position of the ball versus time? Just solve for Vi given that the ball returns to the same y position after 3.2 seconds.

3. Nov 18, 2013

Medgirl314

Thanks berkeman! Are you saying that I should use this equation, but take into account that the acceleration is negative?

4. Nov 18, 2013

Staff: Mentor

Which what where equation?

5. Nov 18, 2013

Medgirl314

Sorry, wrong demonstrative pronoun. *That* equation.

Which is *this* equation:a=Δv/Δt.

6. Nov 18, 2013

Staff: Mentor

Oh, now we're using BIG words!

No, I mentioned the equation for the vertical position of the ball as a function of time. Can you start with that equation?

7. Nov 18, 2013

HallsofIvy

Try using s(t)= -at^2+ vt+ h where a= 9.8 m/s^2, v is the initial velocity and h the initial height. Since you are talking about a ball going up and back down, you can take the initial height to be 0. The quadratic equation, -9.8t^2+ vt=0, has two solutions. One is obviously 0, the time the ball leaves your hand. The other solution is the time the ball returns to your hand which you are told 3.2 seconds. So solve -9.8(3.2)^2+ v(3.2)= 0 for v.

8. Nov 18, 2013

Medgirl314

Haha! You should stick around.

I may be able to, but I haven't gotten into vertical position yet, unless you're talking about the change in height.

9. Nov 18, 2013

Medgirl314

Thank you!!! I will try this and post my answer in a moment.

10. Nov 18, 2013

Medgirl314

983.4496+v(3.2)=0

Do I take the square root of both sides now? Or do I use the quadriatic equation?

11. Nov 18, 2013

Medgirl314

I think I can use the quadriatic equation, but I'm having some trouble figuring out which numbers to assign to the letters.

12. Nov 18, 2013

Staff: Mentor

Please show more of your work. I know that Halls basically solved the problem for you, but you need to do some of this schoolwork on your own.

Show us the definition of the Quadratic Equation. Show us how it works (look at wikipedia, for example). Then show us how you think you should use it to solve this problem.

13. Nov 18, 2013

Medgirl314

Well, I wouldn't have had trouble with that part anyway, I supposse, it's just assigning the letters that seems difficult. Here's the definition I'm looking at: http://www.purplemath.com/modules/quadform.htm

-9.8(3.2)^2+ v(3.2)= 0

Would -9.8 be a, v would be b, and 3.2 would be c?

Thanks!

14. Nov 18, 2013

Medgirl314

Or would -9.8(3.2) be a, v would be b, and 3.2 would be c?

15. Nov 18, 2013

Staff: Mentor

Actually, since the equation does not have a constant term (since the ball returns to the same height), you can just solve the equation directly for Vi. Sorry that I missed seeing that earlier. What answer do you get?

16. Nov 18, 2013

Medgirl314

I'm not quite sure yet. Do I take this:-9.8(3.2)^2+ v(3.2)= 0 and divide by v? Or subtract v?

17. Nov 18, 2013

Staff: Mentor

Start with that equation and take the first term on the lefthand side (LHS) over to the RHS...

18. Nov 18, 2013

Medgirl314

Okay, thanks! I've done this plently of times with binomials and trinomials, but the quadriatics seem a bit harder.
-9.8(3.2)^2+ v(3.2)= 0
v(3.2)+9.8(3.2)=0

Like that?

19. Nov 18, 2013

Staff: Mentor

I'm not following what you did. My suggestion was to take the first term on the LHS over to the RHS. That would put the = sign between the two terms. Then simplify...

20. Nov 18, 2013

Medgirl314

I added the -9.8(3.2)^2 to v(3.2). Oops.

-9.8(3.2)^2+ v(3.2)= 0
v(3.2)=9.8(3.2)

More like that?

21. Nov 18, 2013

Staff: Mentor

It looks like you dropped the ^2 part on the RHS, but that's along the right lines other than that...

22. Nov 18, 2013

Medgirl314

Oops, typo. Thanks! Where did the =0 go?
v(3.2)=9.8(3.2)^2
v(3.2)=885.6576
276.768=885.6576

That doesn't add up. Where did I go wrong?

23. Nov 18, 2013

Staff: Mentor

Something is wrong with your calculator...

On the RHS, square the 3.2, *then* multiply by 9.8.

Then divide the RHS by 3.2 to solve for Vi...

24. Nov 18, 2013

Medgirl314

Okay, thanks! I thought the whole thing was squared.

v(3.2)=9.8(3.2)^2
v(3.2)=9.8(10.24)
v(3.2)=100.352
v=31.36 m/s

25. Nov 18, 2013

Staff: Mentor

Looks reasonable.

Remember, the equation you were starting with was:

$$y(t) = y_0 + v_{iy} * t + \frac{1}{2} * a_y * t^2$$

It's important to keep your terms in mind, so that you don't end up squaring the wrong thing(s).