Probably Simple 1D Kinematics problem

[ccgrad05x2]

Homework Statement

To find the height of an overhead power line, you throw a ball straight upward. The ball passes the line on the way up after 0.70 s, and passes it again on the way down 1.2 s after it was tossed

What is the height of the power line?
m
What is the initial speed of the ball?
m/s


I have just started spring term physics and was doing okay on this homework assignment till i got to this last problem. I am guessing that the force of gravity will play a role, but am stuck at where to start

Thanks so much

 

[Office_Shredder]

There are 1.2 seconds between the first and second time the ball gets tossed into the air. So it takes .6 seconds for the ball to reach the top of its arc after it passes the wire, meaning it takes 1.3 seconds for the ball to reach the top of its arc after throwing it.

Does that help?

 

[ccgrad05x2]

Hmm..they way i understood the problem, the ball is only being tossed once in the air, and passes the wire after .70 s then passes it again .50 s later (1.2). This would mean that the ball was at its peak at .95 s

I could be totally wrong though...this problem just is not clicking with me

 

[hage567]

Hmm..they way i understood the problem, the ball is only being tossed once in the air, and passes the wire after .70 s then passes it again .50 s later (1.2). This would mean that the ball was at its peak at .95 s

I could be totally wrong though...this problem just is not clicking with me

I agree with this.

What equations are you trying to use? Remember that at the maximum height, the velocity is zero. That might help.

 

[ccgrad05x2]

So, this could be way off, if it has a velocity of zero at the top, then the ball would fall at an acceleration rate of 9.8 m/s^2 since only gravity would be acting on it? So would that mean that the maximum height would be just 9.8 x .95 since that was the amount of time that the ball was able to fall. That would give me a height of 9.31 meters, the balls peak, which would then have the height of the power line being 6.86 meters..but I do not think this will work out it doesn't account for the initial velocity of the ball going up.

I feel like I am missing some key part...

 

[hage567]

You're not making sense with your equation. The max height would not be just 9.8*0.95 (check the units, it doesn't work). But yes, one way you can get the height of the wire is by considering the time it takes the ball to fall from the peak. That part doesn't depend on the initial velocity. Do you know an equation relating height (or displacement), time and acceleration? Look at your notes, maybe.

Initial velocity you can get by thinking about the definition of acceleration, using the velocity is zero at the peak fact again.

 

[ccgrad05x2]

Oh!

Maybe got it now:
So the equation would be:
Displacement = (initial velocity)t + .5 at^2
Therefore,
Displacement = 0 + .5(9.8)(.25)
The .25 representing the time the ball had to fall.
So that would make the answer, 1.225 meters.

Thanks you so much for your help hage567. If you wouldn't mind, posting just once more to let me know if this was the right equation, I would greatly appreciate it.

Its so simple now..just couldn't see it.

 

[hage567]

That's the right equation to find the height, but you are not using the right time. You want the time it takes to drop from the max height back to the starting point (your hand), right? Fix that and I think it should be OK.