An athlete executing a long jump leaves the ground at a 29.5

In summary, an athlete leaves the ground at a 29.5∘ angle and travels 7.69m. If this speed were increased by just 4.0%, the jump would be 8.27m.
  • #1
Superstar
2
0

Homework Statement


An athlete executing a long jump leaves the ground at a 29.5∘ angle and travels 7.69m . A- What was the take off speed? B- If this speed were increased by just 4.0%, how much longer would the jump be?

Homework Equations

- Unfortunately I'm totally stuck and I'm not sure what equations to use for this[/B]

The Attempt at a Solution

- I can't come to any solution because I don't really understand where to begin with this problem..[/B]
 
Physics news on Phys.org
  • #2
What you know is:
  • In the jump phase, the only force is g which pulls the athlete down
  • The start angle tells you which part of the take off speed goes upward
  • ... and which part goes forward
  • When the height above ground is 0 (again), the athlete lands
  • And you are supposed to know the formula for distance given speed and accelration
 
  • #3
G**gle SUVAT.
 
  • #4
I'm sorry, but I have no idea what that means. I just started Physics for the first time a few days ago, and am having a very hard time
 
  • #5
Superstar said:
I'm sorry, but I have no idea what that means. I just started Physics for the first time a few days ago, and am having a very hard time
Well, if you had searched the net for SUVAT, as I suggested, you might well have found this, for example: http://en.wikipedia.org/wiki/Equations_of_motion.
Searching for suvat in that page would bring you to http://en.wikipedia.org/wiki/Equations_of_motion#Uniform_acceleration.
However, that reference is not quite ideal because it fails to point out that you can solve a problem such as yours by analysing horizontal and vertical motion separately. That is, you apply the 1-dimensional equations ("collinear" there) in the horizontal and vertical directions separately. To do this, you have to know how to resolve a velocity into its horizontal and vertical components.
There are lots of other hits for SUVAT. See which you find the most intelligible. This will be much more efficient than people on this forum trying to teach it to you. But, if there's any specific statements in them you need to have more explanation on, I'll see what I can do.
You might also find this useful, once you have the basics: https://www.physicsforums.com/insights/frequently-made-errors-mechanics-kinematics/
 

Related to An athlete executing a long jump leaves the ground at a 29.5

1. How is the distance of a long jump measured?

The distance of a long jump is measured from the take-off point to the nearest mark made in the sand by any part of the athlete's body.

2. What is the significance of an athlete leaving the ground at a 29.5" mark in a long jump?

The 29.5" mark indicates the height at which the athlete left the ground during the long jump. This height is used to calculate the athlete's vertical velocity and trajectory, which can be used to analyze their performance.

3. How does an athlete's take-off angle affect their long jump distance?

The take-off angle, or the angle at which the athlete leaves the ground, can greatly impact their long jump distance. A more optimal angle can result in a longer jump, while a less optimal angle can result in a shorter jump.

4. What factors can affect an athlete's ability to jump and reach the 29.5" mark in a long jump?

An athlete's strength, speed, technique, and body mechanics can all play a role in their ability to jump and reach the 29.5" mark in a long jump. Training and conditioning in these areas can help improve an athlete's performance.

5. How do scientists use data from a long jump to study an athlete's performance?

Scientists can analyze data from a long jump, such as take-off angle, velocity, and distance, to study an athlete's performance and identify areas for improvement. This data can also be compared to other athletes to gain insights and inform training programs.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
4K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
24
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
11
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
1K
Back
Top