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jacobwh · Mar 24, 2010, 08:51 AM

Two balls are thrown in the air. One is 10kg and the other is 100kg.

The balls are thrown from the same point, at at the same velocity.

Do both balls reach the same height in the air?

ebaines · Mar 24, 2010, 09:40 AM

jacobwh - what do you think the answer is, and why?

Hint - you will get a different answer depending on whether you consider the effect of air resistance.

jacobwh · Mar 24, 2010, 10:19 AM

Well I am considering this for the effects of a vertical jump. And I think (not considering the effects of air resistance) that they would be the same.

I guess I think that two balls projected at the same velocity will reach the same height in the air.

My reasoning is that although the heavier ball has greater pull of gravity it also has greater inertia to overcome the pull of gravity. The smaller ball has less inertia but less mass and thus less gravity so they both equal out and reach the same height.

Love to hear what you think!

ebaines · Mar 24, 2010, 01:50 PM

You are correct- the larger mass of one ball means it is pulled on harder by gravity, but its larger initial intertia (actually its greater mass) exactly balances that out. The height the ball will reach (neglecting air resistance) is:

h = v^2/2g

Notice that the height is dependent only on the initial velocity v and the acceleration due to gravity, g.

Now for extra credit: what do you think happens if you consider air resistance?

jacobwh · Mar 24, 2010, 02:05 PM

This is great! I've been looking for someone who was smart in this stuff!

The larger ball, considering it was greater in surface area would be more affected by friction and would not go as high?

Ending velocity of a jumper then, will determine the height of the jump, regardless of weight or strength (which may aid in creating velocity)... right? Barring air resistance.

This would also mean, that is we know the jumping height, and the pull of gravity, we can determine starting velocity?

ebaines · Mar 24, 2010, 02:28 PM

For air resistance we must consider the relative surface area per unit volume, because it's surface area that controls air resistance. The small ball has more surface area per unit volume, meaning it has more surface area per unit of mass, and hence it will be more affected by wind resistance, since it has less inertia for the air resistance to overcome. Hence the small ball does not rise as far as the large ball, assuming that they both start with the same initial vertical velocity.

The physics of a human jumping is a little more compicated, but in essence if you tell me the height that his center of gravity rises from the instant his toe leaves the ground I can tell you what his initial vertical velocity was at that instant. Of course, for a large person to have the same initial vertical velocity as a small person does require more strength. On the other hand, a short person has to jump higher to clear the same bar as a tall person (because his center of gravity starts at a lower height off the ground), and hence the short person needs a greater initial vertical velocity. I guess that's why high jumpers and pole vaulters tend to be tall people!

jacobwh · Mar 24, 2010, 03:12 PM

Ahh I see my flaw in the air resistance! Nice! Although greater area for friction it has less friction, per unit of volume, in proportion to it's inertia. Righty?

That's interesting you say that. I was once in a discussion with another trainer, and I insisted that even strength is only valuable to a jumper if it adds to the ending velocity of the jump, because it is THAT which determines the height of a jump.

Because velocity, presupposed that more or less strength is required (depending on the mass) to reach that high velocity.

His argument was that a jumper with the same velocity, but who was also "stronger" and applies more force would jump higher. But more force transferred to an object, could only be manifest in greater velocity. Right?

jacobwh · Mar 24, 2010, 03:15 PM

" if you tell me the height that his center of gravity rises from the instant his toe leaves the ground I can tell you what his initial vertical velocity was at that instant."

Do you know the formula to calculate this? Basically this means that if someone has a 30 inch vertical, we could also say that they have "___" take off velocity. Eh?

ebaines · Mar 25, 2010, 05:52 AM

Originally Posted by jacobwh

His argument was that a jumper with the same velocity, but who was also "stronger" and applies more force would jump higher. But more force transferred to an object, could only be manifest in greater velocity. Right?

Right!

ebaines · Mar 25, 2010, 05:58 AM

Originally Posted by jacobwh

Do you know the formula to calculate this? Basically this means that if someone has a 30 inch vertical, we could also say that they have "___" take off velocity. Eh?

I gave you the formula earlier:

$ h = v^2/2g $

Rearrange and you get:

$ v = \sqrt {2gh} $

For a 30-inch vertical (meaning his center of gravity rises 30 inches from the instant his toes leave the ground) you get:

$ v = \sqrt {2gh} = \sqrt {(2 \times 32.2 \frac {ft} {s^2} \times 30 \ inch \times \frac {1\ ft} {12\ inch} } = \sqrt {161 \frac {ft^2} {s^2} } = 12.7 \frac {ft} s. $

jacobwh · Mar 25, 2010, 12:15 PM

Now that is cool. Thanks alot.

Please stop me if I am dragging this post along, but this is just fascinating stuff.

If an athlete laid on their back, and "pushed" a ball into the air with their feet, it could be an estimation of the type of velocity their legs are capable of creating.

For example if the ball could be pushed into the air 48 inches, then they can contract the muscles at a velocity of 16 ft/second. (Given the form is different then jumping).

So the problem here is that this person is not able to overcome the resistance of their own bodyweight, in order to create the same velocity they were able to create with the ball.

Is it possible create a formula for the amount of force in lbs needed, in a short amount of time, to create a certain velocity?

People come up with power=force X velocity types of formulas as a formula for the vertical jump. But I think it is flawed, because high force and low velocity could be a great amount of power, but if there is not a time constraint, the object or person may not even leave the ground. Is this correct?

BTW - How do you have all these answers at the tip of your tongue Baines? Who are you?

ebaines · Mar 25, 2010, 02:16 PM

Jacobwh - no problem - you're asking some very interesting questions - not the usual "can you do my homework for me" type of question that we often get!

There is indeed a formula that may help here: it relates the impulse applied to an object (defined as the force times the length of time that force is applied) to the change in momentum it experiences:

$ Ft = mv $

Rearrange:
$ F = \frac {mv} t $

Here $F$ is the sum of the force applied by the jumper's muscles (let's call that $F_L$ ) MINUS the opposing force of gravity. So plugging these terms in:

$ F_L = mg + \frac {mv} t $

The trick is trying to understand the value of $t$ , which for a jumper is the length of time from the instant he starts his upward jumping movement to the instant his feet leave the ground. Maybe you have an estimate of that time?

I don't think that power is a relevant quantity here. A more appropriate factor is energy - the energy the athlete expends applying force to the ground as he jumps is equal to the force of his push off times the distance applied, and will be equal to his change in kinetic energy:

$ Fd = \frac 1 2 m v^2 $

Or putting into the same terms as the earlier equation:

$ F = mg + \frac 1 2 \frac {mv^2} d $

So now you have two formulas - one that requires you to know the time duration of applied force, and the other to know the distance of applied force.

As to how do I have these answers - well, let's just say my wife thinks I'm a nerd! I studied engineering in college (many years ago), and have always enjoyed math and physics problems. I must say that if the level of questions gets passed the sophomore physics or math level I am out of my comfort zone!

jacobwh · Mar 25, 2010, 02:57 PM

As I read this it makes me realize that my direction might be wrong here.

I was thinking that a certain "time" was necessary, but now I realize that time isn't the "essential" element. Although, it takes most jumpers about a 1/4 of a second to execute the movement, even if it took longer it's the ending velocity that matters.

Hmmm...

I'll muster that, or maybe you can clarify me, feeling confused.

Are inertia and momentum the same?

So if we had a person whose weight was 200 pounds, and they applied 300 lbs of force, and it took 1/4 of a second to perform, and in that quarter second the waist move from start position to 12 inches upwards (to reach "toe off" or "take off")... then hmmm... I don't know what to do with it!

I want to be able t plug in someone's weight, lbs of force, and estimate the take off velocity it would create? Can I do that with this formula?

Sometimes science just kicks me in the pants!

Well this is a cool site. I didn't realize it was for homework, and I just thought I'd ask some physics questions I've been wrastlin with for a while to see if I could parse them out. And within 15 minutes I was in contact with a "nerd" or more what I call a genius! Haha

ebaines · Mar 26, 2010, 06:51 AM

Let's run the numbers you gave me:

The athlete's weight = 200 lb.
$F_L$ = 300 lb, and is applied in the upward direction over a distance of 12"

So how high will he jump?

From that second equation I gave you earlier you have:

$ F_L = mg + \frac 1 2 \frac {mv^2} d $

The term $mg$ is the athnlete's weight, which we can call $W$ .

Remember we can correlate $v$ to the height he jumps by:

$ v = \sqrt{2 gh}, \\ v^2 = 2gh $

So:
$ F_L = W + \frac 1 2 \frac {m(2gh)} d \\ 300\ lb = 200\ lb + 200\ lb \cdot \frac h {12"} \\ h = 6" $

Your jumper won't exactly make the state championships with that kind of leg strength!

jacobwh · Mar 26, 2010, 08:00 AM

How are we able to estimate the height of the jump without imputting somewhere the element of time it takes to cover the distance? It seems like there is insufficient data to figure out velocity from just having the force, and the body weight...

300 lbs of upward force will overcome the 200lb weight of the jumper, but how do we figure out velocity?

Maybe I am missing something obvious.

jacobwh · Mar 26, 2010, 08:00 AM

How are we able to estimate the height of the jump without imputting somewhere the element of time it takes to cover the distance? It seems like there is insufficient data to figure out velocity from just having the force, and the body weight...

300 lbs of upward force will overcome the 200lb weight of the jumper, but how do we figure out velocity?

Maybe I am missing something obvious.