# Homework Help: Ball thrown upward and elapsed time

1. Sep 19, 2006

### dboy83

A ball is thrown upward from the edge of a building that is 100m tall. On its way back down, the ball just misses the building, falling to the ground below. The elapsed time between the throw and the ball hitting the ground is 6.5 sec.

A) The initial speed of the ball
B) Position and speed of the ball after 1 sec and after 4 secs.
C) The maximum height reached by the ball, measured above the ground.

*Ok, I've worked a few problems where the object is dropped directly from the top of the building but none where the object is actually thrown upward!*

2. Sep 19, 2006

Hint: initial position.

3. Sep 19, 2006

### dboy83

so...

So, would the initial speed be 0m/s or is that only when the ball is at rest?

4. Sep 19, 2006

No, the initial speed does not equal zero. Remember, the ball is thrown. You can get the initial speed directly from the equation which represents the displacement of the ball, x(t) = x0 + v0*t + 0.5*g*t^2. Just set the x-axis along the building and everything will be more clear.

Last edited: Sep 19, 2006
5. Sep 19, 2006

### neutrino

Just x0, not x0*t.

6. Sep 19, 2006

Thanx, I mistyped. :)

7. Sep 19, 2006

### dboy83

when you say, (t) = x0*t + v0*t + 0.5*g*t^2 does that translate to
initial displacement times time, plus initial velocity times time, plus 0.5 times gravity times time squared?

would the initial displacement equal 100 meters? time 6.5 sec?
so, I just plug and chug.....

8. Sep 19, 2006

Once again, I apologise for my mistype. x(t) = x0 + v0*t + 0.5*g*t^2. Yes, that translates to initial displacement (100 m) plus velocity times time etc. The time is 6.5 seconds. BUT, there is one more condition. The left side of the equation has to equal zero, since at that time the ball fell on the ground, and we have set the coordinate axis x so that the origin (x=0) is at the bottom of the building (the ground), and x = 100 m is the point from which we threw the ball.

9. Sep 19, 2006

### dboy83

so, 0= 100m + v0*6.5s + 0.5*-9.80m/s^2*6.5s^2 ? then I solve for the initial velocity?

10. Sep 19, 2006

Yes, exactly. I'm glad you noticed g must be negative.

11. Sep 19, 2006

### dboy83

hmmm

ok, I have 0=100m + v0 * 6.5s + 0.5 * -9.80m * 6.5 after I cancel out the s^2. I'm not sure how I add the meters and seconds..... shouldn't I have seconds in the demoninator? Would I use pemdas rule to add these?

12. Sep 19, 2006

What? You don't have to add any meters nor seconds. You're dealing with numbers only.

13. Sep 19, 2006

### dboy83

I have 100 m and 6.5 s. I'm confused now...

14. Sep 19, 2006

Yes, and you have all you need. You already wrote down the equation you need to solve. Again, 0 = 100 + v0*6.5 + 0.5*(-9.81)*6.5^2 .

15. Sep 19, 2006

### dboy83

So, how should I go about multiplying and adding the numbers? I tried to multiply and add the numbers out but i got -106.525, which can't be the answer...

16. Sep 19, 2006

### dboy83

i think i have it

0 = 100 + v0*6.5 + 0.5*(-9.81)*6.5^2
= 100 + v0*6.5 + (-207.025)
= -107.025 + v0*6.5
= 107.025= v0*6.5
= 16.5= v0
so, 16.5 m/s =v0 ?

17. Sep 19, 2006

Yes, looks good. I'm sure you could do this by yourself (the last step), and I don't think such steps need to be discussed here.

18. Sep 19, 2006

### dboy83

where did you..

where did you get the formula from? where does the 0.5 come from in the formula?

19. Sep 19, 2006

These are basic kinematic equations which can be found everywhere, and it would be weird of your teacher to assign problems to you for which you do not posess the knowledge to solve; I believe he mentioned these equations at some point.

20. Sep 19, 2006

### dboy83

thanks

I couldn't find it anywhere in my book. I still don't understand where the 0.5 came from. I really want to know how physics works instead of plugging and chugging. Thanks for your help. Without brainiacs like you, I would be so lost..

21. Sep 19, 2006

To be honest, I'm no brainiac, sadly. Well, the internet is just the right place to start learning physics. I suggest you start googling up things like 'equations of motion', 'basic kinematics', etc.

22. Sep 19, 2006

### dboy83

formula for b?

what formula would I use for B?

23. Sep 19, 2006

Use the formula we already mentioned and used: x(t) = x0 + v0*t + 0.5*g*t^2. The reason why we 'called' it x(t) is exactly the reason of interest in B) - every time value we 'plug into' this equation will give us the position at that certain moment. So, x(t) means that the position x of the ball depends of the time t. For example, for t = 1 sec, we have x(1) = x0 + v0*1 + 0.5*g*1^2 = something.. (of course, you have to plug in the values of x0, v0, and g, too) I hope you got the point.

For the speed, use the equation v(t) = v0 + g*t.

24. Sep 19, 2006

### physicsgal

how good are you at algebra? if you notice in some of the acceleration/displacement formulas, there's a "2" in them. so if you switch the formula around to solve for a different variable, you might end up with a "1/2" = 0.5.

~Amy

25. Sep 20, 2006

### dboy83

displacment negative or positive?

For part B, I found the positions to be 11.6m for 1 sec and -12.4m for 4 sec. I'm not sure if I'm supposed to take into consideration the height of the building and add it to those positions, or just state that those are the positions relative to the release point. I believe the question is asking for the position relative to the release point since part c is more specific and asks for the position relative to the ground. What do you think? Also, for part c, the maximum height is 13.9m relative to the release point but it asks for the height relative to the ground. So, do I add 100m + 13.9m to get the max height relative to the ground? It seems logical to just add the two to get the max height, but the part that confused me is the displacement of the building is -100m not +100m. For some reason, this part confuses me because I thought I was supposed to add the displacment of the building, which is -100m , plus the displacement of the ball from the release point, which is 13.9m, to get the max height.....