# 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