Diver Time & Velocity Question

In summary, the diver will be in the air for 1.39 seconds and have a velocity of 13.65 meters per second when diving off of a 10 meter spring board.
  • #1
x2017
84
1

Homework Statement



Consider the Olympic sport of diving. The more time the athlete spends in the air, the greater the potential to complete difficult maneuvers. Off of a 3m spring board with a stiffness of 5920N/m, a diver with a mass of 52kg will maximally deflect the board 0.5m from it's equilibrium position. Assume that all of the strain energy stored in the boards returned to the diver in the vertical direction.
a) How much time will the diver be in the air when using the 3m spring board?
b) What will the diver's velocity be when they hit the water?
c) How much time will a diver spend in the air when diving off of a 10m platform?
d) What will the diver's velocity be when they hit the water?

Homework Equations


t=(2d/g)1/2
KE=1/2mv2
SE=1/2kΔx2

The Attempt at a Solution


[/B]
Sorry for putting a photo in, if you can't make it out let me know and I'll type it out!
IMG_0244.jpg


Since the velocities ended up being the same I think I may have gone about it the wrong way...
 
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  • #2
When does your first relevant equation apply ? Can you use it for this exercise ?
b) and d) look very much the same to me. Would you get the same answer for a 30 m board too :smile: ?
 
  • #3
BvU said:
When does your first relevant equation apply ? Can you use it for this exercise ?
b) and d) look very much the same to me. Would you get the same answer for a 30 m board too :smile: ?

I figured when the diver jumps they're like a projectile so that's why I used that time equation.
Yes, if I continued to use that method, but logically I know that that wouldn't be the case I just am unsure how else to go about it.
 
  • #4
<<Moderator note: Response to deleted post removed>>
So for c) I did this:
t=(2h/g)1/2
t=[(2)(9.5)/(9.81)1/2
t=1.39s

Used a height of 9.5m instead of 10m because the board was deflected 0.5m.

Aaaaaaand, this for d)
v=(2gh)1/2
v=[2(9.81)(9.5)]1/2
v=13.65m/s

Used a height of 9.5m instead of 10m because the board was deflected 0.5m.
 
Last edited by a moderator:
  • #5
Using initial height of 9.5m doesn't make sense. One, because the platform doesn't deform like the spring board. Two, because that would be assuming that the diver is simply falling off the deformed board without taking advantage of any upward bounce.
What you calculated for the strain energy from the spring board looks to be correct, I would assume that velocity would be reached when the board returns to the 3m point where max velocity occurs, not at the 2.5m point where velocity is clearly equal to zero.
 
  • #6
RUber said:
Using initial height of 9.5m doesn't make sense. One, because the platform doesn't deform like the spring board. Two, because that would be assuming that the diver is simply falling off the deformed board without taking advantage of any upward bounce.
What you calculated for the strain energy from the spring board looks to be correct, I would assume that velocity would be reached when the board returns to the 3m point where max velocity occurs, not at the 2.5m point where velocity is clearly equal to zero.

I didn't notice the change in wording, thanks for pointing it out!
 
  • #7
As a former teacher of H.S. physics, I find that the rules of this forum actually prevent what I consider to be "help".
So having unintentionally violated, and not being in agreement with the one rule that states "no solutions allowed", I will no longer participate.
 
  • #8
Jim Kadel said:
As a former teacher of H.S. physics, I find that the rules of this forum actually prevent what I consider to be "help".
So having unintentionally violated, and not being in agreement with the one rule that states "no solutions allowed", I will no longer participate.
Giving a solution to someone is just like giving them a fish but not teaching them how to fish. If you teach your students through solutions only without making them understand the concept by them self, They won't solve most of the questions.

I am a H.S student and that is what I believe.

Sorry for going out of the topic. Keep trying to solve OP until you succeed and you will feel the achievement of solving something by your own
 

1. What is the formula for calculating diver time and velocity?

The formula for calculating diver time and velocity is velocity = distance / time. This means that the velocity of the diver can be calculated by dividing the distance traveled by the time it took to travel that distance.

2. How do you measure the time and velocity of a diver?

The time and velocity of a diver can be measured using various methods such as a stopwatch, underwater cameras, or sensors. The most accurate way to measure these parameters is by using specialized equipment such as a dive computer or a velocity meter.

3. What factors can affect the time and velocity of a diver?

There are several factors that can affect the time and velocity of a diver. These include the diver's physical fitness and technique, water temperature, currents, and the weight of their equipment. Additionally, changes in air pressure and depth can also impact the time and velocity of a diver.

4. How can you improve your time and velocity as a diver?

To improve your time and velocity as a diver, you can focus on improving your technique and physical fitness. Proper breathing techniques, streamlined body positioning, and efficient use of equipment can also help increase your speed and reduce your overall time in the water.

5. Why is it important to track diver time and velocity?

Tracking diver time and velocity is essential for safety and performance purposes. It allows divers to monitor their progress, identify areas for improvement, and ensure they are following safe diving practices. It also helps dive professionals and researchers gather data and better understand the capabilities of divers in different environments.

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