Physics Problem

[JaceTheAce]

Alright, I need help on this one... can't figure out which equation to use. How do you get the answer?

A 200 lb. circus performer jumps from a 20 foot ladder onto a teeter-totter, launching a 50 pound child. Assume a perfect transfer of energy, with no friction or air resistance. In his arc of travel, the child lands 50 feet horizontally to the rear. What was the total height the child achieved?

Answer: 75.9 feet

[I_Money]

Are you taking a calculus based class? Or are we assuming a is constant? And why are you in the English system, and not SI - my first step regardless is to convert lbs to Slugs.

[TXaviator]

hmm, wouldnt the object launched travel straight up if launched from a teeter -totter? what is imparting the horizontal force?

[I_Money]

It was not obivously launched straight up, which means you will have to work out V for X/Y.

[JaceTheAce]

Originally Posted by I_Money

Are you taking a calculus based class? Or are we assuming a is constant? And why are you in the English system, and not SI - my first step regardless is to convert lbs to Slugs.

Yes, I think we're supposed to convert to Slugs.

[TXaviator]

well, in essence, arent you just going to find the force (m*a) applied to the launcher, then since you know the force and mass of the child launched, that gives you his acceleration....and you know that he will then be decelerated by g.....giving you the time in air....

this is much easier to explain if i hadnt had a six pack (cooled by my brand new EXPRESSJETS coozie i found around campus yesterday!)

[I_Money]

TX that is assuming acceleration is constant - BIG ASSUMPTION!! That is why we need to know if he is a calc based class!!!

If it is an alg. class you can not work out Y unless you know Vy, how are we going to work that out with only knowing X.

[TXaviator]

youre right. i only had alg based physics!! i succumb to the lords of mechanics.

[JaceTheAce]

Originally Posted by TXaviator

well, in essence, arent you just going to find the force (m*a) applied to the launcher, then since you know the force and mass of the child launched, that gives you his acceleration....and you know that he will then be decelerated by g.....giving you the time in air....

this is much easier to explain if i hadnt had a six pack (cooled by my brand new EXPRESSJETS coozie i found around campus yesterday!)

But what would the max height be? We already know the answer, but how is it calculated?

[Pilot Hopeful]

If g is not constant, well, that is sadistic. However, for a 20-foot difference, we can assume g is nearly constant. Thus, the equations are indeed greatly simplified.

How about starting with energy conservation? Assuming total conservation of mechanical energy, the initial gravitational potential energy of the performer with respect to the surface (mgh) must be equal to his kinetic energy as he lands on the teeter-totter (1/2 mv^2). The only unknown variable is v. Solving for v, we can then determine his momentum, p = mv.

Next, we assume a total energy transfer from performer to baby, i.e.,
E(performer) = E(baby)
(1/2)m(perf)(v(perf)^2) = (1/2)m(baby)(v(baby)^2)

Solving this equation we can determine v(baby). Note that this v(baby) may have x and y components.

Finally, we must solve kinematics equations to find max baby height.

[I_Money]

G will be constant - unless the circus went to the moon.

[fish314]

Here is how you go about this problem. I'm going to number my paragraphs below to make it easire to refer back to them later...


Step 1. Solve for the man's potential energy.

First you need to figure out how much energy was imparted to the teeter-totter from the man's jump. The formula for potential energy is P.E.=mgh. So convert the man's weight to slugs (that gives you m), g is 32.2 ft/s^2 and the height is 20 ft. That potential energy will be converted to kinetic energy as the man falls. That's energy due to his motion.


Step 2. Solve for the total initial velocity of the baby. (Vitotal)

Since the problem assumes no elastic effects (perfect transfer of energy) all of that energy will be transfered to the 50 lbs. child. So all of the man's potential energy get's turned into kinetic energy and all of that kinetic energy gets transfered to the baby. P.E. = K.E.

The 50 lbs. child was launched both up and backwards away from the teeter-totter, right? You know this because the problem tells you that he landed 50 feet horizontally to the rear of the teeter-totter. So you're next step is going to be figuring out how much of the energy was UP and how much was Back ( Motion in the Y direction and motion in the X direction). Start by figuring out what the total intial velocity of the baby was. You get that from the kinetic energy equation. K.E.=.5mV^2. You already know what the K.E. is. It's your answer from the P.E. Equation above, because all of that potential energy from the man was transfered to the baby. The m in this case is the weight of the baby, 50lbs, converted to slugs. Then you solve for V. I'm going to refer to this V from now on as Vitotal (Initial velocity TOTAL) This V is not in either the straight up direction or the straight out direction, it's a combination of both (hence TOTAL). But it's your starting point for the next section of work.

Step 3. Solve for initial velocity in X and Y (Vix and Viy)
Ok, at this point, you know a couple of things about the baby's motion. You know his total initial velocity (but you don't yet know the x or y component of the total), you know his acceleration in the y direction (which is just gravity: 32.2 ft/sec^2 in the DOWN direction or -32.2 ft/sec^2 in the UP direction) and you know his total distance traveled in the x diretion, 50 ft. You also know that there is basically no acceleration in the x direction once the baby gets moving, so you just handle that by assuming an initial velocity in the x direction, and assuming that the baby starts moving instantaneously. Don't worry too much about this assumption, all we are saying is that there is no acceleration in the x direction.

Now there are 4 kinematic equations that we are going to use here, but separate them into the 2 directions that we are concerned with, x and y. Here are the 4 equations:

D =Vi*t + .5 a*t^2 (distance travelled= initial velocity multiplied by time plus half of the acceleration multiplied by time squared)

Vf^2=Vi^2+2*a*d (final velocity squared = initial velocity squared plus 2 times acceleration times distance travelled)

Vf=Vi +a*t (Final velocity=initial velocity plus acceleration *time)

And last but not least:

D= .5(Vf+Vi)*t (Distance = 1/2 of final velocity plus initial velocity times time)


Step 3a. Solve for time, t, in terms of Vix
If you use this last equation, you could solve for the total time that the baby was airborne, in terms of the velocity in the x direction. This is because you know that there was no acceleration in the x direction so therefore vf and vi in the x direction are the same. So the time is equal to D/Vix (where Vix denotes the initial velocity in the x direction, which we still don't know). Or time, t, equals 50ft/Vx.

Step 3b. Solve for Viy in terms of Vix
The next thing we are going to do is to find initial velocity in the Y direction in terms of Vx initial. To do this we are going to use the equation Vf=Vi +a*t. Now we know how much time it takes for the baby to go all the way up, and then come all the way back down, at least in terms of initial velocity in X. That's 50ft/Vx. But we are going to use a trick....

What happens when you throw a baby (or anything else) up in the air. It starts out going up, and it slows down, then it stops at the top of the arc, then it speeds up coming down. At the top of the arc, the velocity in the Y direction is 0, and that occurs exactly half-way through the throw, time wise.

So, if we substitute into the equation above, and we are looking at the first half of the throw:

Vfy=Viy+ay*t. In this case, Vfy (final y velocity) is the velocity at the top of the arc, or 0. Acceleration in the y direction is -32.2 ft/sec^2, and the t that we are interested in is HALF of the total time the baby is airborne or .5*50ft/Vix. We now can solve for Viy (initial y velocity) in terms of Vix. Viy=32.2*.5*50/Vix.


Step 3c. Solve for Vix and Viy using Vitotal
Ok. Next step is to solve for Vix (initial velocity in x). We know total initial velocity from above. We solved for it when we solved the kinetic energy equation. I'm not sure if you are familiar with how you calculate the magnitude (size) of a vector from it's components, but here is how.

Vitotal^2=Viy^2 +Vix^2. Since we know Vitotal and we know Viy in terms of Vix, we can solve for Vix:

Vix^2 +(.5*32.2*50/Vix)^2=Vitotal^2.

Once you've solved for Vix, then solve for Viy:

Viy=(.5*32.2*50/Vix)

And while we're at it, we may as well solve for the time, t.

t=50ft/Vix (that's for the whole flight)

and t=25ft/Vix (the time it takes for the baby to get to the top of the arc, or half of the total flight).


Step 4. Solve for the height.

Ok, go back to those kinematic equations. Remember this one:
D =Vi*t + .5 a*t^2? Well now, using what we know we can solve for the distance that the baby travels in the UP direction, which is how high the baby gets. We just need to put in all the appropriate terms for just the Y direction. I'll replace the letter D (distance) with the letter H, for Height, but they mean the same thing.

H=Viy*t +.5 ay *t^2... Ok. We've already solved for Viy. The t that you are looking for is the half flight time or 25ft/Vix. And ay is just -32.2.



And once you do all that work, you should come up with the right answer. PM me if you still have questions.

[Cruise]

Originally Posted by fish314

Here is how you go about this problem. I'm going to number my paragraphs below to make it easire to refer back to them later...


Step 1. Solve for the man's potential energy.

First you need to figure out how much energy was imparted to the teeter-totter from the man's jump. The formula for potential energy is P.E.=mgh. So convert the man's weight to slugs (that gives you m), g is 32.2 ft/s^2 and the height is 20 ft. That potential energy will be converted to kinetic energy as the man falls. That's energy due to his motion.


Step 2. Solve for the total initial velocity of the baby. (Vitotal)

Since the problem assumes no elastic effects (perfect transfer of energy) all of that energy will be transfered to the 50 lbs. child. So all of the man's potential energy get's turned into kinetic energy and all of that kinetic energy gets transfered to the baby. P.E. = K.E.

The 50 lbs. child was launched both up and backwards away from the teeter-totter, right? You know this because the problem tells you that he landed 50 feet horizontally to the rear of the teeter-totter. So you're next step is going to be figuring out how much of the energy was UP and how much was Back ( Motion in the Y direction and motion in the X direction). Start by figuring out what the total intial velocity of the baby was. You get that from the kinetic energy equation. K.E.=.5mV^2. You already know what the K.E. is. It's your answer from the P.E. Equation above, because all of that potential energy from the man was transfered to the baby. The m in this case is the weight of the baby, 50lbs, converted to slugs. Then you solve for V. I'm going to refer to this V from now on as Vitotal (Initial velocity TOTAL) This V is not in either the straight up direction or the straight out direction, it's a combination of both (hence TOTAL). But it's your starting point for the next section of work.

Step 3. Solve for initial velocity in X and Y (Vix and Viy)
Ok, at this point, you know a couple of things about the baby's motion. You know his total initial velocity (but you don't yet know the x or y component of the total), you know his acceleration in the y direction (which is just gravity: 32.2 ft/sec^2 in the DOWN direction or -32.2 ft/sec^2 in the UP direction) and you know his total distance traveled in the x diretion, 50 ft. You also know that there is basically no acceleration in the x direction once the baby gets moving, so you just handle that by assuming an initial velocity in the x direction, and assuming that the baby starts moving instantaneously. Don't worry too much about this assumption, all we are saying is that there is no acceleration in the x direction.

Now there are 4 kinematic equations that we are going to use here, but separate them into the 2 directions that we are concerned with, x and y. Here are the 4 equations:

D =Vi*t + .5 a*t^2 (distance travelled= initial velocity multiplied by time plus half of the acceleration multiplied by time squared)

Vf^2=Vi^2+2*a*d (final velocity squared = initial velocity squared plus 2 times acceleration times distance travelled)

Vf=Vi +a*t (Final velocity=initial velocity plus acceleration *time)

And last but not least:

D= .5(Vf+Vi)*t (Distance = 1/2 of final velocity plus initial velocity times time)


Step 3a. Solve for time, t, in terms of Vix
If you use this last equation, you could solve for the total time that the baby was airborne, in terms of the velocity in the x direction. This is because you know that there was no acceleration in the x direction so therefore vf and vi in the x direction are the same. So the time is equal to D/Vix (where Vix denotes the initial velocity in the x direction, which we still don't know). Or time, t, equals 50ft/Vx.

Step 3b. Solve for Viy in terms of Vix
The next thing we are going to do is to find initial velocity in the Y direction in terms of Vx initial. To do this we are going to use the equation Vf=Vi +a*t. Now we know how much time it takes for the baby to go all the way up, and then come all the way back down, at least in terms of initial velocity in X. That's 50ft/Vx. But we are going to use a trick....

What happens when you throw a baby (or anything else) up in the air. It starts out going up, and it slows down, then it stops at the top of the arc, then it speeds up coming down. At the top of the arc, the velocity in the Y direction is 0, and that occurs exactly half-way through the throw, time wise.

So, if we substitute into the equation above, and we are looking at the first half of the throw:

Vfy=Viy+ay*t. In this case, Vfy (final y velocity) is the velocity at the top of the arc, or 0. Acceleration in the y direction is -32.2 ft/sec^2, and the t that we are interested in is HALF of the total time the baby is airborne or .5*50ft/Vix. We now can solve for Viy (initial y velocity) in terms of Vix. Viy=32.2*.5*50/Vix.


Step 3c. Solve for Vix and Viy using Vitotal
Ok. Next step is to solve for Vix (initial velocity in x). We know total initial velocity from above. We solved for it when we solved the kinetic energy equation. I'm not sure if you are familiar with how you calculate the magnitude (size) of a vector from it's components, but here is how.

Vitotal^2=Viy^2 +Vix^2. Since we know Vitotal and we know Viy in terms of Vix, we can solve for Vix:

Vix^2 +(.5*32.2*50/Vix)^2=Vitotal^2.

Once you've solved for Vix, then solve for Viy:

Viy=(.5*32.2*50/Vix)

And while we're at it, we may as well solve for the time, t.

t=50ft/Vix (that's for the whole flight)

and t=25ft/Vix (the time it takes for the baby to get to the top of the arc, or half of the total flight).


Step 4. Solve for the height.

Ok, go back to those kinematic equations. Remember this one:
D =Vi*t + .5 a*t^2? Well now, using what we know we can solve for the distance that the baby travels in the UP direction, which is how high the baby gets. We just need to put in all the appropriate terms for just the Y direction. I'll replace the letter D (distance) with the letter H, for Height, but they mean the same thing.

H=Viy*t +.5 ay *t^2... Ok. We've already solved for Viy. The t that you are looking for is the half flight time or 25ft/Vix. And ay is just -32.2.



And once you do all that work, you should come up with the right answer. PM me if you still have questions.

LOL....nice explanation!

[BobDDuck]

Wow... It's been 4 years since I touched a physics or calc book but scarily enough I was still able to follow that. Granted I couldn't have come up with any of it, but give a blind man a fishing pole.... wait... that's not right.


Nice write up fish.

[CapnJim]

Wait a minute...where did you get an ExpressJet coozie?

[TFaudree_ERAU]

Thanks Jace, you just helped me remember how much I don't miss that chit.

[Persian_Pilot]

Hmmm I have no idea what you all are talking about but ,

JaceTheAce, Who's that chick with you? :

[TXaviator]

Originally Posted by CapnJim

Wait a minute...where did you get an ExpressJet coozie?

was laying around Odegard Hall unattended with some other expressjet schwag.... the coozie was a no-brainer *yoink*

[CapnJim]

Score!

[Cruise]

Originally Posted by Persian_Pilot

JaceTheAce, Who's that chick with you? :

It doesn't really matter. They are ALL annoying and a P.I.T.A.

Says a jaded male.....

[TXaviator]

Originally Posted by Cruise

It doesn't really matter. They are ALL annoying and a P.I.T.A.

Says a jaded male.....

sounds like someone isn't getting any lately