Physics - The proverbial ball being thrown, both up and down, and where do they meet?

[64th]

So the question is horribly vague and we were told to, by any means, figure it out. I'm begging all of you, to put me out of my misery here.

 

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A ball is thrown up into the air at an upward velocity of 25m/s, consequently, a ball is thrown down towards the ground from a building 15 meters high. At what point do the two balls meet, and (I'm assuming here) at what hieght?

 

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Again, no details were given in reference to the force thrown, or displacement. Say the ball dropped from the building has an initial velocity of 0m/s to make things a bit easier. And say we don't take into consideration how tall the person throwing the ball is. That it is, mysteriously, just moving off of the ground on it's own accord.

 

Any help given would be GREATLY appreciated, as this stupid question has killed my weekend. (Though, the second of the two problems is pretty much solved).

[NeonBlack]

I'm begging all of you, to put me out of my misery here.

 

*bang* You're in a better place now, Old Yeller.

 

Hint: Let the people have 0 height.

 

What do you think you need to do? What progress have you made in the problem? What have you figured out so far?

[Cap'n Refsmmat]

You'll need to write equations for both pieces of motion. Try doing that before going further.

[64th]

Oh god. u.u I'll try as best I can to organize my thoughts then.

 

Problem TWO: The guy throwing the ball down from 15 meters.

Okay, easy. I have a handy formula for this. Find the 'knowns':

 

vi: 0m/s

vf: ?

a: +9.81m/s^2

d: 15m

t: ?

 

t= + sqrt[0-2(9.81m/s^2)(15m)]/9.81m/s^2

t= 1.75s

 

vf= vi + at

vf= 0 + (9.81m/s^2)(1.75s)

vf= 17.17m/s

 

Then I did a little chart of it, manually, detailing the displacement after every .25s:

 

.25s = .318825 m

.50s = 1.226525 m

.75s = 3.67875 m (I don't think this is correct)

1.00s = 4.905 m

1.25s = 7.6640625 m

1.50s = 11.03625 m

1.75s = 15.0215625 m

 

So that problem checked out okay. I'll be really heartbroken if it's wrong.

[Cap'n Refsmmat]

The trouble is that you want to find at what height the balls meet. So you'll need an equation for both that can tell you their heights at any given time.

[64th]

Problem ONE: The Evil One, The Throwing Up One.

So, I realize there are actually two steps to this one. Finding all the unknowns for the peak and then going back down again. With THAT said...

 

Step I -

 

vi: 25m/s

vf: 0m/s

a: -9.81m/s^2

d: ?

t: ?

 

So, I tried two things with T. The first equation I used initially in Problem TWO, and came up with:

 

t: 0.3924s

 

And then I did it again, this time with...

 

t= vf - vi / a

t= -25m/s / -9.81m/s^2

t= 2.5484s

 

Okay. What? Wait... No no... No. D:

 

And yet, I proceed with the first equation.

 

d= di + vi(t) + 1/2 at^2

d= 0 + (25m/s)(.3924s) + 1/2 (-9.81m/s^2)(.3924s^2)

d= 9.81 - .755

d= 9.055m

 

And then with the second.

 

d= di + vi(t) + 1/2 at^2

d= 0 + 25m/s(2.55s) + 1/2(-9.81m/s^2)(2.55s^2)

d= 63.75 - 31.89475

d= 31.855m

 

D: Is it possible for a ball to travel upwards 31 meters in under 3 seconds? Somehow.... I have my reservations.

 

Right. So afterwards I proceeded to turn myself around in circles. If you really want the confused math, I guess I could post it all. But mostly, if you guys can provide me with a formula that will give me a decent answer... I can finish it off.

 

The trouble is that you want to find at what height the balls meet. So you'll need an equation for both that can tell you their heights at any given time.

 

I believe I used the freefall equation thing-

 

d: .5(9.81m/s^2)(t^2)

 

To do the chart.

 

Or would that be cheating? Probably... Google isn't helping much either. :/

[timo]

Preface: The follwing might seem a bit too complicated to some. But I think following the path 64th took to solving the problem is a good idea, pedagogically.

 

Preface 2: The equation of motion for an object with constant acceleration is [math] \vec x(t) = \vec x_0 + \vec v_0 t + \frac{1}{2} \vec a t^2 [/math] with [math]\vec x(t)[/math] being the position at time t, [math]\vec x_0[/math] being the initial position (the position at time 0), [math]\vec v_0[/math] being the initial velocity (guess at which time ) and [math]\vec a[/math] being the acceleration. This equation, in conjunction with [math] \vec v(t) = \vec v_0 + \vec a t [/math], is all the physics you need to know/remember for those types of questions about motion with constant acceleration (note that "no acceleration" is also a constant acceleration) - the rest is rearranging equations and using rational thinking.

 

64th, I've not read any of your equations, they seem a bit confusing to me. Anyways:

Then I did a little chart of it, manually, detailing the displacement after every .25s

Let's follow this approach. It's the pedestrian version but leads to a result and is potentially easy to understand:

Draw those charts into a graph (position over time) for both balls. At the position where the two lines (where with "line" i mean "curve", they are not straight lines) intersect, both balls have the same position at the same time -> that's exactly what you are looking for if I'm not mistaken. This graphical way of solving the problem definitely is a solution to the problem. If you want a more elegant solution, then think about how to convert that way into mathematical expressions:

- Can you express the lines in the graphs in a mathematical way? Hint: Yes, of course. I wouldn't propose it if it wouldn't work. Judging from the graph you drew it should be appearant that expressing both lines via a function of position as a function of time seems like a good approach.

- Hopefully having expressions for both lines, the next question is how "the lines intersect in one point" translates to a mathematical expression. Well, I've already mentioned the mathematical expression for the two intersecting lines: They have the same value X for position at the same time T, i.e. position_of_ball_1(at time T) = position_of_ball_2(at time T).

- This equation can then be solved for the time T at which they meet. Depending on the number of additional assumptions you put into your initial conditions, T will be a function of some number of variables (those variables being exactly those initial conditions that you assumed no value for). For a start I strongly recommend fixing all of the initial conditions you mentioned as not being given to some (ideally sensible) value, e.g. the initial velocity of the upper ball being zero and the lower ball starting at a height of zero.

[Fred56]

Don't you need a function that defines the distance between the two objects? Then you solve for when it's = 0?

The hint from the problem about the 15 stories means ball two can probably be assumed to travel at terminal velocity...

[timo]

Don't you need a function that defines the distance between the two objects? Then you solve for when it's = 0?

Assuming that was in reply to my post. Yes. But: Define the distance d between the two balls with position p1 and p2 as d(t) = |p1(t) - p2(t)|. Then, your condition is equivalent to setting both positions equal, i.e. d(t) = |p1(t)-p2(t)|=0 <=> p1(t) = p2(t).

 

The hint from the problem about the 15 stories means ball two can probably be assumed to travel at terminal velocity...

I see no hint about 15 stories (what size are the people in your area ). The ball travels exactly at the velocity that it travels at, as given by the equations in my previous post. There is no need to use approximations here (except ignoring air resistance), I don't even think it's justified doing so.

[Fred56]

consequently, a ball is thrown down towards the ground from a building 15 meters high.

Sorry, I read the problem wrong. But since it says the ball is thrown, presumably it then has some undefined velocity? Wouldn't it just be easier to assume it will reach vt before reaching the point of intersection? Unless it really is a trick question.

 

Nah, that's a bunch of horseshite. 15 meters is way too small a distance to reach terminal v. So the only other assumption (that I think makes any sense) is that ball 2 is "dropped" from a height of 15m. That sounds more like it. Maybe I shouldn't have had that smoke.

[Country Boy]

Since the problem specifically says "thrown down", I would be inclined to think the initial downward speed of the second ball is the same as the initial upward speed of the first ball, 25 m/s. I would also ignore air resistance. The acceleration for the two balls is the same: 9.8 m/s^2 downward.

 

Taking the ground to be y= 0, at time t seconds, the first ball will have height y1(t)= -4.9t^2+ 25t. The second ball will have height y2(t)= -4.9t^2- 25t+ 15. They will pass when those heights are the same: -4.9t^2+ 25t= -4.9t^2- 25t+ 15. Since the "-4.9t^2" terms will cancel, that's a simple linear eqaution to solve for t. Once you have found t, evaluate either y1 or y2 (or both, as a check) for the height of the two balls.