Velocity of Pole Vaulter: Solving for 10.95 m/s

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A pole vaulter reaches a height of 6 meters, and the discussion focuses on calculating the velocity upon hitting the ground using the equation mgh = 1/2mv^2, which simplifies to v = √(2gh). The calculated velocity of approximately 10.95 m/s is confirmed as correct by multiple participants. An alternative method using the kinematic equation v^2 = 2gh also yields the same result, demonstrating consistency in the calculations. The conversation touches on the importance of understanding energy conservation in physics, while also humorously addressing the need for regular class attendance. Overall, the methods discussed effectively solve the problem of determining the pole vaulter's impact velocity.
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Homework Statement



A pole vaulter goes 6m in the air. What is the velocity when he hits the ground?

No air resistance, just g=9.8m/s^2

Homework Equations


mgh = 1/2mv^2

Does an answer of 10.95 seem right? (I used 10m/s/s for gravity)
 
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Well, then you need another approach, maybe attending class more regularly?

I'm not being snide but for the life of me cannot think of any other method.

The one option which might not be considered strictly as kinematics is by using energy: the potential and kinetic energies under these conditions will be constant. So develop an eqn that uses both. But given the data, this is the only approach.
 
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denverdoc said:
Well, then you need another approach, maybe attending class more regularly?

I'm not being snide but for the life of me cannot think of any other method.

The one option which might not be considered strictly as kinematics is by using energy: the potential and kinetic energies under these conditions will be constant. So develop an eqn that uses both. But given the data, this is the only approach.

Attending class would be wonderful...last week was beyond my control

anyways, not to go off on a tangent, using mgh=1/2mv^2, I canceled out mass, since it's just a multiplier to both, and got an answer of approximately 10.95. Does that seem about right?
 
whiteruskii said:
Attending class would be wonderful...last week was beyond my control

anyways, not to go off on a tangent, using mgh=1/2mv^2, I canceled out mass, since it's just a multiplier to both, and got an answer of approximately 10.95. Does that seem about right?
If you use an alternate approach (and yes, there is one) using the basic kinematic equation of motion
v^2 = 2gh, you arrive at the same result (same equation), even without a knowledge of energy methods. Don't forget the units for the velocity: m/s
 
Yes that sounds about right. BTW, the other method would be to use the kinematic equation for distance, with the initial velocity = 0, and figure out how long it takes to travel the 6m distance in free-fall, then use that time in the velocity kinematic equation to figure out the person's velocity at that time. But your energy approach is better, because it's all contained in one easy equation.

BTW, are you guys classmates? Sounds like you have a conversation going on behind the scenes...
 
Doh! PhantomJay beat me to the punch.
 
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