Olympic Long Jumper's Takeoff Speed

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In summary, the Olympic long jumper takes off at an angle of 21.8 ° and jumps 8.97 m horizontally before landing. The takeoff speed is found to be 8.97m/s.
  • #1
makdaddymac
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Homework Statement


An Olympic long jumper leaves the ground at an angle of 21.8 ° and travels through the air for a horizontal distance of 8.97 m before landing. What is the takeoff speed of the jumper?

Homework Equations


Square root of V^2_ox + V^2_oy

The Attempt at a Solution



Sqrt of 8.97^2 + (-9.8)2, idk what to do with the 28degree
 
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  • #2
Your equation to combine the components of velocity doesn't exactly apply to the data given.

The problem gives you the distance of the jump. Not the horizontal component of V.
 
  • #3
This is a projectile motion problem. You have to separate the vertical and horizontal motions because different formulas apply. The horizontal part is motion at constant speed while the vertical part is accelerated.

Use Vx = v*(21.8), Vy = v*sin(21.8)

For the horizontal part, use x = Vx*t
For the vertical part, use V = Vy + at and y = Vy*t + .5at^2
Put the numbers you know in all three formulas. It should then be possible to solve one of them and find something out (usually the time) so you substitute in another of them and find what you want.
 
  • #4
makdaddymac said:

Homework Statement


An Olympic long jumper leaves the ground at an angle of 21.8 ° and travels through the air for a horizontal distance of 8.97 m before landing. What is the takeoff speed of the jumper?

Homework Equations


Square root of V^2_ox + V^2_oy

The Attempt at a Solution



Sqrt of 8.97^2 + (-9.8)2, idk what to do with the 28degree
Let [tex]O(X,Y)[/tex] be a frame of reference with origin in the jumper. The equations of velocity and space in function of time are:

Velocity:

[tex]
v_x=v_o cos\theta
[/tex]

and

[tex]
v_y=v_o sin\theta + g*t
[/tex]

Space

[tex]
s_x=v_o cos\theta * t +s_{ox}
[/tex]

and

[tex]
s_y=\frac{1}{2}g*t^2 + v_o * sin \theta + s_{oy}
[/tex]

I assume [tex]s_{ox}[/tex] and [tex]s_{oy}[/tex] null inasmuch as origin is in the jumper. So the relation [tex]s_y=f(s_x) [/tex] by deleting [tex]t[/tex] is:

[tex]s_y=\frac{1}{2}g*\frac{s_x^2}{(v_o * sin \theta)^2} + s_x * tan\theta} [/tex]

By substituting [tex] \theta=21.8°[/tex], [tex]s_x=8.97m [/tex]and [tex]s_y=0 [/tex] you'll get [tex]v_o[/tex]I hope this help, bye...:wink:
 
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Related to Olympic Long Jumper's Takeoff Speed

1. What is the average takeoff speed of an Olympic long jumper?

The average takeoff speed of an Olympic long jumper is approximately 9-10 meters per second, or about 22 miles per hour.

2. How is the takeoff speed of a long jumper measured?

The takeoff speed of a long jumper is typically measured using a specialized instrument called a speed gun, which uses radar technology to accurately record the athlete's speed.

3. What factors can affect the takeoff speed of a long jumper?

The primary factors that can affect the takeoff speed of a long jumper include their technique, physical fitness, and the surface of the runway. Wind speed and direction can also play a role in their takeoff speed.

4. How does the takeoff speed of a long jumper affect their overall performance?

The takeoff speed of a long jumper is a crucial factor in their overall performance. A faster takeoff speed can give the athlete more momentum and a longer jump distance. It also allows them to have a more efficient takeoff angle, which can improve their landing position.

5. Can the takeoff speed of a long jumper be improved?

Yes, the takeoff speed of a long jumper can be improved through proper training and technique. Strength and power training, as well as refining their takeoff technique, can help increase their speed. Additionally, working on their reaction time and explosiveness can also contribute to a faster takeoff speed.

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