Hard physics homework problem

[yhelothar]

This is a problem based on motion. (velocity, acceleration, etc).

Abby throws a ball straight up and times it. She sees that the ball goes by the top of a flagpole after 0.500s and reaches the level of the top of the pole again after a total elapsed time of 4.10s. What was the speed of the ball at launch?

So how am I supposed to figure this out while only knowing the time? We don't know at what distance from the ground the ball was launched, we don't know how tall is the flagpole.

 

[silverpig]

Assume the ball started from 0.

And yeah I could figure it out, but it's your homework, and I already have mine to do

 

[yhelothar]

Originally posted by: silverpig
Assume the ball started from 0.

And yeah I could figure it out, but it's your homework, and I already have mine to do

Even if the ball started from 0, I have no idea how to figure it out either.
The flag pole could be 100m tall or 5m tall, you don't know.

Actually I don't even quite get what's happening in the question. It says it reaches the top of the flagpole again, which implies that it bounced up again? Huh? Even if it bounced up again, it would not reach its original height. This question makes no sense to me.

 

[DaShen]

HINT: treat the top of the flag pole as zero and work out the 4.10 s first. Since it is going straight up and down, this should be easy.

Once you get that velocity, use the .5 s to work that velocity back to the original velocity.

It has been a while since Mechanical Physics, so I can't help you with equations much. Plus it is your homework.

**EDIT**
Remember 2.05 s is when the ball crests from the flagpole. So it should be easy to figure out the drop.

(9.8 m/s^2) * 2.05 s = velocity at point of flag.

 

[yhelothar]

Oh I get it now.. yeah it's pretty easy actually.
D'oh!

 

[silverpig]

The height of the flagpole will be one of the last things you figure out. And the question means that the ball is thrown up, passes the top of the pole, continues climbing, stops, falls back down and passes the top of the pole again. Break the problem up into 3 parts (before reaching top of pole, between times it reaches top of pole, and after it passes the top on the way back down). Solve the middle section to get an initial velocity, then use that initial velocity as the final velocity for section 1.

 

[Jittos]

oh boy, i haven't done any physics since my first year in college, which is a looong ass time ago.
Anyways, I prob don't remember enough to help u solve the problem. but here's my thought.

so it takes 2.3 seconds for the ball to go from start to the top, right? u should be able to calculate that distance given G = 9.8 something.
once u have the total distnace, u calculate the speed at the start as if u drop the ball from that height.

I think this is do-able. gluck

 

[Howard]

It goes up past the top, then it falls back down to the same point at 4.10s.

You can calculate the height of the flagpole given the velocity of the ball right after the throw or vice versa. Since you're missing that piece of info, solve for it using the 2nd time given.

 

[DaShen]

(2.55 s <to crest overall> ) * 9.8 m/s^2

.5 + 4.1/2 for crest time.

 

[BehindEnemyLines]

Think of the ball going up once to the top in 0.500s then falls downward and reaches the top level again at 4.10s. You know the the time and acceleration. Unless the problem is wrong (I doubt that), I'm sure you can find one or two equations from your physics book that relate time, acceleration (in your case gravity at 9.81m/s^2), and velocity.

 

[yhelothar]

I get it now. DaShen's advice to treat the top of the flagpole at 0 made it make sense.
Basically it takes 4.1 seconds to fall to the ground, meaning 4.1 times the acceleration of gravity would give me the final velocity, and thus the height of the flagpole.

 

[yhelothar]

Originally posted by: DaShen
(2.55 s <to crest overall> ) * 9.8 m/s^2

.5 + 4.1/2 for crest time.

Wait.. I see that the crest to ground time isn't 4.1s now. So where did you get the 2.55s crest time?

 

[HN]

i get 45m/s but i ain't none too good at them there things

 

[NoMoMoney]

.5 seconds to pole - 1.8 seconds to crest - 1.8 seconds to pole (total elapse time of 4.1) - .5 seconds to hit spot it was launched at

So you would use 2.3 to crest.

 

[yhelothar]

Originally posted by: HN
i get 45m/s but i ain't none too good at them there things

That's one of the multiple choice answers.
So I'm assuming you got it right.
How did you figure out the time it took for the ball to fall to the ground from the crest of the flagpole?

 

[HN]

Originally posted by: virtualgames0

Originally posted by: HN
i get 45m/s but i ain't none too good at them there things

That's one of the multiple choice answers.
So I'm assuming you got it right.
How did you figure out the time it took for the ball to fall to the ground from the crest of the flagpole?

wait, is 24.99 also one of the answers? i was about to revise.

 

[ahurtt]

The actual height of the flagpole doesn't really matter very much if at all. What your physics teacher is trying to get you to realize is that you can solve the question by knowing that the gravitational acceleration constant is 9.8 meters per second per second. For extra credit you could tell your teacher how tall the flagpole was.

 

[DaShen]

4.1 s / 2 = 2.05 s //you must half the time to get the crest time
.5 s to get to flag pole

2.05 + .5 = 2.55 s to get to crest 2.55 s (no friction and no wind resistance)

2.55 s * 9.8.... m/s^2 = final answer.

clear enough?

 

[NoMoMoney]

Originally posted by: DaShen
4.1 s / 2 = 2.05 s //you must half the time to get the crest time
.5 s to get to flag pole

2.05 + .5 = 2.55 s to get to crest 2.55 s (no friction and no wind resistance)

2.55 s * 9.8.... m/s^2 = final answer.

clear enough?

You have to add .5 to go from flag pole to point of origin (4.1 is total elapsed time, not time since passing the flagpole) on the way back down, so
(4.1 + .5)/2 = 2.3

 

[HN]

physics....it's grrrrrrrrrrrrrrrrreeeeeeeat

 

[DaShen]

Originally posted by: NoMoMoney

Originally posted by: DaShen
4.1 s / 2 = 2.05 s //you must half the time to get the crest time
.5 s to get to flag pole

2.05 + .5 = 2.55 s to get to crest 2.55 s (no friction and no wind resistance)

2.55 s * 9.8.... m/s^2 = final answer.

clear enough?

You have to add .5 to go from flag pole to point of origin (4.1 is total elapsed time, not time since passing the flagpole) on the way back down, so
(4.1 + .5)/2 = 2.3

She sees that the ball goes by the top of a flagpole after 0.500s and reaches the level of the top of the pole again after a total elapsed time of 4.10s.

If the total elapsed time it 4.10 from when the ball was thrown to when it comes back to the flag pole and not the elapsed time between reaching the flag pole and coming back to the flag pole, then the equation is

(4.1 - .5)/2 + .5 = 2.3

(4.1 - .5) is the elapsed time between poles in this case half that is crest time between poles. Add .5 for the total crest time.

otherwise if it is the time elapsed from in between the flag pole times then 2.55

2.3 s * 9.8 m/s^2 = answer if this is the case. which is 22.54

You might need to clarify this.

 

[yhelothar]

Originally posted by: DaShen
4.1 s / 2 = 2.05 s //you must half the time to get the crest time
.5 s to get to flag pole

2.05 + .5 = 2.55 s to get to crest 2.55 s (no friction and no wind resistance)

2.55 s * 9.8.... m/s^2 = final answer.

clear enough?

So we must assume that the ball was thrown from ground point to solve this problem right?

 

[NoMoMoney]

Doesn't matter where the ball was thrown from. The key to figuring out the problem is knowing that in ideal conditions (ignoring friction) that the travel time up is equal to the travel time down. With ideal conditions the velocity at any height of x going up equals the velocity at height x coming down. I throw the ball from my hand and it takes 2.3 (or 2.55) seconds to reach the highest point (at which point it is not moving). If I keep my hand in the same position, 2.3/2.55 seconds later it will hit my hand with the same velocity as I threw it. I can throw it from the ground, on a chair, on top of a car, or even standing on your shoulders.