# A ball is thrown vertically upward

1. Aug 6, 2009

### bumblebeeliz

two objects thrown in the air. HELP!

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

A ball is thrown vertically upward with an initial speed of 11 m/s. One second
later, a stone is thrown vertically upward with an initial speed of 25 m/s. (a) Find
the time it takes the stone to catch up with the ball. (b) Find the velocities of the
stone and the ball when they are at the same height.

vo= 11 m/s
vo=25 m/s
v = ?
a/g = -9.80 m/s2
y = ?
yo = 0
t = ?

2. Relevant equations

y= yo + vo*t + 1/2 at^2

3. The attempt at a solution

Ball Vo= 11m/s
Stone Vo= 25m/s

I have chosen the y=yo+vo*t+1/2 at^2 kinematic equation:

Ball: y= 0 + 11m/s * t + 1/2 (-9.80 m/s2)*t^2

Stone: y= 0 + 25m/s * t + 1/2 (-9.80 m/s2) *t^2

Then I use the Ball=Stone to find the time between the two which equals to:

11m/s * t = 25m/s * t

But this does not get me anywhere. I thought I would eventually use the quadratic formula, but nothing led me to it. I always seem to have trouble when there are two things moving at different times and speeds. Any tips or tricks?

Last edited: Aug 6, 2009
2. Aug 6, 2009

### prob_solv

Re: two objects thrown in the air. HELP!

See the quote, Feel the quote, Understand the quote, enlightened ?

3. Aug 6, 2009

### kuruman

You are not taking into account the fact that the stone was thrown one second later. Imagine a clock attached to each object. The clock starts running when motion starts. This means that the stone's clock (when it starts moving) will always read one second less than the ball's clock. Do you see now how you must change the equation describing the motion of the stone?

4. Aug 6, 2009

### bumblebeeliz

Thanks Kuruman.

Does this mean that the time (t) for the stone is -1? And therefore equals to:

Stone: y= 0 + 25m/s * (-1) + 1/2 (-9.80 m/s2) *(-1)^2

I understand what you mean by "the stones clock will always read one second less than the balls clock" but I am really not sure how to include it in my equation. I wish I could see this problem animated!

5. Aug 6, 2009

### rl.bhat

Before the stone catch the ball, the ball must be moving in the air for longer duration than the stone by one second.
Displacement for both must be the same.

6. Aug 6, 2009

### kuruman

No. It means that if by t you mean "the time recorded by the ball's clock" then t-1 is the time recorded by the stone's clock. t-1 is always one second behind t, whatever t happens to be.

7. Aug 6, 2009

### bumblebeeliz

Ok. So if I recap my variables:

BALL:

vo= 11 m/s
v = ?
a/g = -9.80 m/s2
y = ?
yo = 0
Bt = t

y= 0 + 11m/s * t + 1/2 (-9.80 m/s2)*t^2

STONE:

vo= 25m/s
v = ?
a/g = -9.80 m/s2
y = ?
yo = 0
St = t-1s

y= 0 + 25m/s * t-1 + 1/2 (-9.80 m/s2)*t-1^2

makes more sense?

8. Aug 6, 2009

### kuruman

Yes. Now solve for the catch-up time and don't forget the parentheses as in (t-1) or (t-1)2.

9. Aug 6, 2009

### bumblebeeliz

I tried several ways of calculating the two equations together and I always seem to end up with the same problem:

11 m/s (t) + ½(-9.80 m/s2) (t)2 = 25 m/s (t-1) + ½ (-9.80 m/s2) * (t-1)2
11 m/s (t) – 4.9 m/s2 (t)2 = 25 m/s * (t-1) – 4.9m/s2 * (t-1)2
11 m/s (t) – 4.9 m/s2 (t)2 = 25 m/s t – 25 m/s – 4.9 m/s2 t2 + 4.9 m/s2
0 = (25 m/s t – 11 m/s t) – (25 m/s) – (4.9 m/s2 t2 + 4.9 m/s2 t2) + 4.9 m/s2
0 = 14 m/s t – 25 m/s + 4.9 m/s2

ax2 + bx + c
Am I not missing the “ax2” ?

10. Aug 6, 2009

### kuruman

The correct expansion is

(t-1)2=t2-2t+12

11. Aug 6, 2009

### bumblebeeliz

Second try

11 m/s (t) + ½(-9.80 m/s2) (t)2 = 25 m/s (t-1) + ½ (-9.80 m/s2) * (t-1)2
11 m/s (t) – 4.9 m/s2 * (t)2 = 25 m/s t – 25 m/s – 4.9 m/s2 * (t-1)(t-1)
11 m/s (t) – 4.9 m/s2 * (t)2 = 25 m/s t – 25 m/s – 4.9 m/s2 * (t^2- 2t +1)
11 m/s (t) – 4.9 m/s2 * (t)2 = 25 m/s t – 25 m/s –4.9 m/s2 t2 + 9.8 m/s2 t –4.9 m/s2
0 = 25 m/s t – 11 m/s (t) – 25 m/s + 9.8 m/s2 t – 4.9 m/s2
0 = 14 m/s (t) - 25 m/s + 9.8 m/s2 t – 4.9 m/s2

Again I do not have the values for the quadratic formula so I try factoring and taking out the "t":

0 = 14 m/s (t) - 25 m/s + 9.8 m/s2 t – 4.9 m/s2 (divide all by) / m/s
0 = 14 t – 25 + 9.8 1/s t – 4.9 1/s
-14 t – 9.8 1/s t = -25 -4.9 1/s
t (-14 – 9.8 1.s) = -25 -4.9 1/s
t = -25 -4.9 1/s (divide by) / -14 – 9.8 1/s (but that cancels out the seconds)

this is a tough one!

12. Aug 6, 2009

### kuruman

Writing the equations as you have done with units is fine. (I prefer working with symbols and plug in the numbers at the very end). However the top equation should have terms like (t - 1 s). Number 1 here is not a dimensionless number. Then 25 m/s (t - 1 s) becomes 25m/s t - 1 m. Also, the powers of t2 drop out so you don't really need the quadratic formula.

13. Aug 6, 2009

### bumblebeeliz

wow yeah I didn't catch that. But shouldn't 25 m/s (t-1s) become 25 m/s t -25 m?

I think the answer is: 0.87 s.

Last edited: Aug 6, 2009
14. Aug 6, 2009

### kuruman

Yes. It should be 25 m/s t - 25 m. Check your answer again. At t=0.87 s the stone has not been launched yet.

15. Aug 6, 2009

### bumblebeeliz

Alright. I hope this is right:

0 = 23.8 m/s t - 25 m - 4.9 m/s
-23.8 m/s t = -25m - 4.9 m/s
t= -25m - 4.9 m/s / -23.8 m/s
t= 1.25 s.

16. Aug 7, 2009

### kuruman

The top equation should be
0 = 23.8 m/s t - 25 m - 4.9 m

Check it out. Remember to write (t - 1s). Then

(t - 1 s)(t - 1 s) = t2 - 2(t)(1 s) + 12 s2 = t2 - 2 t s - 1 s2

This method of keeping track of units although correct, is cumbersome and fraught with pitfalls as you have seen. I recommend that you practice using symbols and substituting numbers at the very end. In this case, I would do the following

g = 9.8 m/s2
vB = 11 m/s

Then the ball's height above ground at any time t would be

xB(t) = vBt - ½gt2

For the stone

vS = 25 m/s
t0 = 1 s

and the stone's height above ground at any time t (as measured by the ball's clock) is
xS = vS(t - t0) - ½g(t - t0)2

Then say that there is a specific "catch up" time, tC at which the ball and the stone are at the same height above ground, namely

xB(tC) = xS(tC) or

vBtC - ½gtC2 = vS(tC - t0) - ½g(tC - t0)2

Then do some algebra to solve for tC and finally put in the numbers with their units and make sure that all comes out in the wash.

If it is any consolation, this is probably one of the most complicated problems in linear kinematics that one sees in an introductory course, but it teaches how to handle motion that starts at different times.

17. Aug 7, 2009

### bumblebeeliz

Thanks for taking the time to explain the symbols concept. Sadly, I feel like I get more confused with symbols. But that's perhaps because I need practice...in everything.

I just wrote out the symbols method and I was left over with this:

vb tc - 1/2 gtc^2 = vs tc - vs to - 1/2 g tc^2 + g tc to - 1/2 g to^2

When I only see alphabetical letters, my brain seems to freeze. I'm not sure how to take out the "tc". Am I at the part where we replace the symbols with numbers and units?

The consolation did help!!!

Last edited: Aug 7, 2009
18. Aug 7, 2009

### RoyalCat

Write everything very clearly. Make sure you're able to tell your subscript and superscript apart from your regular script, make sure you can tell your indexes apart. I really can't stress this enough.
Working parametrically can be daunting, but it is also very rewarding, both in terms of results and insights.

If it makes it easier for you to tell $$t_0$$ and $$t_c$$ apart, define a new variable.

$$z\equiv t_c$$

Another useful trick of this sort is to use different letters for the initial velocities of both objects, $$u$$ and $$v$$ are popular choices. Though something you should get used to is using numerical indexes as well, just so your eyes get used to it. You'll find that skill useful once you start dealing with questions that have multiple masses indexed as $$m_1$$,$$m_2$$,$$m_3$$ etcetera (You'll run out of letters eventually!)

And then isolate it in the form of a quadratic equation of the form $$Az^2+bz+c=0$$ where $$a\, b\, c$$ are known quantities.
From there, it should simply be a matter of applying the quadratic root formula.

19. Aug 7, 2009

### bumblebeeliz

Thank you so much for the tricks! I will keep that in mind.

I dont think the quadratic formula is needed in this problem as Kuruman suggested.

20. Aug 7, 2009

### RoyalCat

Ah, yes, sorry, I sort of skimmed over your actual equation since it was hurting my eyes a bit. ^^; Didn't notice your free term was 0, averting the need for a quadratic equation.