I hope this helps.How high did this athlete jump?

In summary: Is there a way to solve this for x without using the equation?Is... Is there a way to solve this for x without using the equation?
  • #1
cy19861126
69
0
The "force platform" is a tool that is used to analyze the performance of athletes by measuring the vertical force as a function of time that the athlete exerts on the ground in performing various activities. A simplified force vs time graph for an athlete performing a standing high jump is shown below (see attachment). The athlete started the jump at t = 0.0s. How high did this athlete jump?

My work so far:
okie, using the equation, I = p:
Ft = mvf - mvi
At the highest point, vf = 0, so...
1000N * 1.0s = -mvi
Now what do I do? I got an equation with two unknowns. I know that once you solve for vi, you can just use the equation: x = vit + 0.5at^2 to solve for x. The question is how do I get vi?

If you can't see the attachment, go to: http://students.washington.edu/cy1126/Physics.jpg
 

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  • #2
There is more information in the graph than you are seeing. What is going on before t = 0? The impulse is the area under the curve of the graph of the net force vs time.
 
  • #3
OlderDan said:
There is more information in the graph than you are seeing. What is going on before t = 0? The impulse is the area under the curve of the graph of the net force vs time.
ohhh, so you're saying that the impulse is actually 0.4N * 1.5s = 0.6Ns? So actually the equation is 0.6 = -mvi. So now what do I do next?
 
  • #4
You can find the mass of the athelete from the graph before 0 sec. Then, plug and solve for v_i.
 
  • #5
lotrgreengrapes7926 said:
You can find the mass of the athelete from the graph before 0 sec. Then, plug and solve for v_i.
Hmm... I don't see what ur saying. How can u get Vi from the graph before 0 sec. Am I missing some concept here?
 
  • #6
w=gm

Weight is in N, g is gravity, m is in kg.
 
  • #7
Alright, I think I'm getting this. So:
w = mg
600N/9.8N/kg = 61.2Kg

Ft = mvf - mvi
Ft = -mvi
(1.5s * 400N * .5) * 1.5s = -61.2kg * Vi (in this part, do you have to divide the force by two like I did since the area under the curve is roughly a right triangle?)
Vi = -7.35 m/s

Therefore, using the equation x = Vit + 0.5at^2
x = -7.35 * 1.5 + 0.5 * -9.8 * 1.5^2 = -22.05m

The answer doesn't seem to be right. I mean at least the sign should be positive
 
  • #8
cy19861126 said:
Alright, I think I'm getting this. So:
w = mg
600N/9.8N/kg = 61.2Kg

Ft = mvf - mvi
Ft = -mvi
(1.5s * 400N * .5) * 1.5s = -61.2kg * Vi (in this part, do you have to divide the force by two like I did since the area under the curve is roughly a right triangle?)
Vi = -7.35 m/s

Therefore, using the equation x = Vit + 0.5at^2
x = -7.35 * 1.5 + 0.5 * -9.8 * 1.5^2 = -22.05m

The answer doesn't seem to be right. I mean at least the sign should be positive
In this equation

Ft = mvf - mvi

vi is zero and vf is the velocity when he leaves the floor at the end of the impulse. vf is positive. After that it is a vertical projectile problem. Your Ft is too big. The 400N and .5 parts are good, but the time base of the triangle is not right. Not only that, it seems you have used the time twice. Ft is the area of the triangle. Check the dimensions of your equation. There can be only one time factor on the left.

You might want to use the equation that relates the change in velocity squared to the acceleration and the distance for the flight (or conservation of energy which is the same thing).
 
  • #9
OlderDan said:
In this equation

Ft = mvf - mvi

vi is zero and vf is the velocity when he leaves the floor at the end of the impulse. vf is positive. After that it is a vertical projectile problem. Your Ft is too big. The 400N and .5 parts are good, but the time base of the triangle is not right. Not only that, it seems you have used the time twice. Ft is the area of the triangle. Check the dimensions of your equation. There can be only one time factor on the left.

You might want to use the equation that relates the change in velocity squared to the acceleration and the distance for the flight (or conservation of energy which is the same thing).

okay, I think I kinda know what ur saying. I'll give it an another shot
Ft = mvf
400N * 1.0s * 0.5 = 61.2 kg * vf
vf = 3.26 m/s

x = vit + 0.5at^2
x = 3.26 * 1s + 0.5*9.8*1^2
= 8.16 m

Is what I am doing correct? I don't get what u said about "You might want to use the equation that relates the change in velocity squared to the acceleration and the distance for the flight (or conservation of energy which is the same thing)." Sorry about this, but I am really bad on this Impulse thing (I was doing perfectly fine on the other conservation of mometum thingy)
 
  • #10
cy19861126 said:
okay, I think I kinda know what ur saying. I'll give it an another shot
Ft = mvf
400N * 1.0s * 0.5 = 61.2 kg * vf
vf = 3.26 m/s

x = vit + 0.5at^2
x = 3.26 * 1s + 0.5*9.8*1^2
= 8.16 m

Is what I am doing correct? I don't get what u said about "You might want to use the equation that relates the change in velocity squared to the acceleration and the distance for the flight (or conservation of energy which is the same thing)." Sorry about this, but I am really bad on this Impulse thing (I was doing perfectly fine on the other conservation of mometum thingy)
I still think your time for the implulse calculation is a bit too long. If you extend the sloping line down to the .6kN line it intersects at about .3sec, so the base of the triangle is about .7sec. The time in the air has no direct connection to the impulse time. It depends only on the velocity achieved at the end of the impulse. The equation that will tell you the maximum height is

v² = v_o² + 2as

At maximum height the velocity is zero, the initial velocity is what you caclulate at the end of the impulse, and a is -g. This equation is equivalent to energy conservation

½mv² + mgh = ½mv_o²
 

Related to I hope this helps.How high did this athlete jump?

1. What is momentum in physics?

Momentum is a measure of an object's motion and is calculated as the product of its mass and velocity. It is a vector quantity, meaning it has both magnitude and direction.

2. What is the formula for calculating momentum?

The formula for momentum is p = m x v, where p is momentum, m is mass, and v is velocity. The SI unit for momentum is kilogram-meter per second (kg*m/s).

3. How is momentum conserved in a closed system?

In a closed system, the total momentum remains constant, meaning it is conserved. This is known as the Law of Conservation of Momentum. This means that the total momentum before a collision or interaction is equal to the total momentum after the collision or interaction.

4. How does mass and velocity affect momentum?

The greater the mass and velocity of an object, the greater its momentum will be. This means that a heavier object moving at a faster speed will have a greater momentum than a lighter object moving at a slower speed.

5. What is the difference between momentum and kinetic energy?

Momentum and kinetic energy are both measures of motion, but they are different quantities. Momentum is a vector quantity, while kinetic energy is a scalar quantity. Additionally, momentum depends on both mass and velocity, while kinetic energy only depends on mass and speed.

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