Jumping from Roof: Find Velocity & Force Exerted

  • Thread starter PearlyD
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In summary, the problem involves a person jumping from a 4.5m high roof and decelerating over a distance of 0.70m. With a torso mass of 45Kg, the questions ask for the velocity just before the feet strike the ground and the average force exerted on the torso by the legs during deceleration. By equating potential energy with kinetic energy and using the equation V^2=2(a)*d, the velocity can be calculated. The average force can then be calculated using the equation f=ma. The force does not come from air resistance, but rather from the stiffness of the muscles in the legs.
  • #1
PearlyD
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Homework Statement


A person jumps from the roof of a house 4.5m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70m. This the mass of his torso(excluding legs) is 45Kg,find (a) his velocity just before his feet strike the ground and (b) the average force exerted on his torso by his legs during deceleration.


2. Relevant Questions
I really don't know how to start this question and all i really got was Force of gravity
Fg=441N
and i don't even know if its used in this question.
 
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  • #2
this is very similar to the other problem you posted, only in reverse.

Since it asks for velocity before he hits the ground--how can we calculate this?

We just equate the potential energy on the roof with the kinetic energy as he strikes the ground:

mgy=1/2mV^2. We can proceed using the approach of the last problem where

F the force of deceleration x distance(d) is equal to mgy (or 1/2mv^2 since they are the same), or using kinematics and looking only at velocity:

V^2=2(a)*d where d is the 0.7m.
 
  • #3
denverdoc said:
this is very similar to the other problem you posted, only in reverse.

Since it asks for velocity before he hits the ground--how can we calculate this?

We just equate the potential energy on the roof with the kinetic energy as he strikes the ground:

mgy=1/2mV^2. We can proceed using the approach of the last problem where

F the force of deceleration x distance(d) is equal to mgy (or 1/2mv^2 since they are the same), or using kinematics and looking only at velocity:

V^2=2(a)*d where d is the 0.7m.

what does mgy mean?
 
  • #4
PearlyD said:
what does mgy mean?

Sorry mgy=mass*gravity*displacement above ground.
 
  • #5
denverdoc said:
Sorry mgy=mass*gravity*displacement above ground.

thanks:)
i got his velocity,but now i need the average force exerted on his torso by his legs during decelertation
Would A =9.8?
I thought it would go like this
Fg-Ar=ma
Where Fg=441 and mass= 45 and acceleration=9.8
but that just gives me a Air resistance of 1 and i don't think that's right
 
  • #6
it is not air resistance which is slowing his descent--it is from the stiffness of the springs (muscles) in his legs--go out and jump off a 3 foot wall. How do you land? With your legs slightly bent and then allow the angle in the knees to decrease so he is in a near squat when his center of mass comes to rest--this is not the same as his feet coming to rest. If this isn't clear, go out and try it!

So what we need is a deceleration that will bring his body (center of mass) to rest in 0.7 meters--distance his chest drops as he "lands".

2a*y=V^2 where v is the velocity at which he hits the ground. Solve for a.

Since f=ma, calculate force.

EDIT: y= 0.7 m if this isn't clear, not the distance off the roof.
 
  • #7
denverdoc said:
it is not air resistance which is slowing his descent--it is from the stiffness of the springs (muscles) in his legs--go out and jump off a 3 foot wall. How do you land? With your legs slightly bent and then allow the angle in the knees to decrease so he is in a near squat when his center of mass comes to rest--this is not the same as his feet coming to rest. If this isn't clear, go out and try it!

So what we need is a deceleration that will bring his body (center of mass) to rest in 0.7 meters--distance his chest drops as he "lands".

2a*y=V^2 where v is the velocity at which he hits the ground. Solve for a.

Since f=ma, calculate force.

EDIT: y= 0.7 m if this isn't clear, not the distance off the roof.

I now got it!
THANK YOU SOOO MUCH!
you do not know how much i appreciated you helping me through this!
Thank you!
 
  • #8
PearlyD said:
I now got it!
THANK YOU SOOO MUCH!
you do not know how much i appreciated you helping me through this!
Thank you!

Cool, and you are most welcome.
 

Related to Jumping from Roof: Find Velocity & Force Exerted

1. How do you find the velocity of a person jumping from a roof?

To find the velocity of a person jumping from a roof, you would need to use the equation v^2 = u^2 + 2as, where v is the final velocity (which is equal to 0 in this case), u is the initial velocity (which is equal to the person's running speed before jumping), a is the acceleration due to gravity (which is approximately 9.8 m/s^2), and s is the height of the roof. Solving for u, you can find the initial velocity of the person jumping from the roof.

2. What factors affect the velocity of a person jumping from a roof?

The velocity of a person jumping from a roof is affected by their initial running speed, the height of the roof, and the acceleration due to gravity. Other factors such as air resistance and the person's body mass may also have a small impact on the velocity.

3. How can you calculate the force exerted by a person jumping from a roof?

To calculate the force exerted by a person jumping from a roof, you can use the equation F = ma, where F is the force, m is the mass of the person, and a is the acceleration due to gravity. The mass of the person can be estimated based on their body weight and the acceleration due to gravity is a known constant. By plugging in these values, you can find the force exerted by the person.

4. Can a person adjust their velocity while jumping from a roof?

Yes, a person can adjust their velocity while jumping from a roof by changing their initial running speed or by using their body to manipulate air resistance. For example, if a person wants to jump with a lower velocity, they can slow down their running speed or spread their arms and legs to increase air resistance. However, it is important to note that the height of the roof and the acceleration due to gravity will also impact the person's velocity.

5. Is it safe to jump from a roof?

Jumping from a roof can be very dangerous and should not be attempted without proper training and safety precautions. The force exerted on the body upon landing can be significant and can lead to injuries. It is always best to consult with a professional or use a safe jumping platform if you are interested in jumping from a roof.

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