Swimmer off tower: Velocity and Force

So the water must push the swimmer up for this to work. You can't push something up by hitting it down.In summary, a 65kg swimmer jumps off a 10.0m tower and reaches a velocity of 14m/s before coming to a stop 2.0m below the surface. The net force exerted by the water on the swimmer is 2,548N in the downward direction. This force is needed to oppose the swimmer's downward velocity and bring them to a stop. The force of gravity also plays a role in the net force, but is not the only force acting on the swimmer.
  • #1
Wanting to Learn
Problem:
A 65kg swimmer jumps off a 10.0m tower.
a. Find the swimmer's velocity on hitting the water.
b. The swimmer comes to a stop 2.0m below the surface. Find the net force exerted by the water.
Given & Equations:
v0 = 0m/s
a = 9.8m/s2
d = 10m
vf = __m/s

F = __N

vf2 = v02 + 2ad
F = ma

Work:
vf2 = v02 + 2ad
v2 = 0 + 2•9.8•10
v2 = 196
v = 14

F = m•a
F = 65•9.8
F = 137.2

Conclusion:
So I got
a. = 14m/s
b. = 137.2N

Is this correct? I just started learning force so have no idea whether that seems to be a reasonable amount of force.
Thank you in advance.
 
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  • #2
Part b is not correct. The swimmer is no longer in free fall therefore his/her acceleration is not 9.8 m/s2. You need to find what acceleration is needed to change the velocity from 14 m/s to zero over 2 meters.
 
  • #3
So would acceleration be -7m/s ?
In that case I would get F = 7*14 = 98, but what would I do with the negative, or is it irrelevant?
 
  • #4
Wanting to Learn said:
So would acceleration be -7m/s ?
How did you get this number? BTW, acceleration has units m/s2.
Also, assuming that a = 7 m/s2, why is F = 7*14? What equation are you thinking of?
The negative is not irrelevant; it denotes a direction opposite to the one you assumed is positive.

On edit: As far as having an intuitive feeling for force is concerned, 1 Newton is approximately the weight of a 1/4 lb. stick of butter.
 
Last edited:
  • #5
Wanting to Learn said:
Find the net force exerted by the water.
The question ought to tell you to pretend that the force is constant during the deceleration, otherwise there is no way to solve it with the information provided.
 
  • #6
Wanting to Learn said:
So would acceleration be -7m/s ?
That is the acceleration required to go from 14 m/s to 0m/s in two seconds. The question at hand is how much acceleration to go from 14 m/s to 0 m/s in two meters.
 
  • #7
I looked in the textbook, found a similar problem, and followed its format (yes I know I should have done that before but because we do not normally use the textbook I hadn't thought to), so I got the following (continued from my earlier work):
vf2 = v02 + 2a(x-x0)
0 = 142 + 2a(2-0)
-196 = 4a
a = -49m/s2

F = m•a
F = 65•-49
F = -3185N

Is this correct?
 
  • #8
kuruman said:
As far as having an intuitive feeling for force is concerned, 1 Newton is approximately the weight of a 1/4 lb. stick of butter.
Thank you for this.
 
  • #9
Wanting to Learn said:
I looked in the textbook, found a similar problem, and followed its format (yes I know I should have done that before but because we do not normally use the textbook I hadn't thought to), so I got the following (continued from my earlier work):
vf2 = v02 + 2a(x-x0)
0 = 142 + 2a(2-0)
-196 = 4a
a = -49m/s2

F = m•a
F = 65•-49
F = -3185N

Is this correct?
Almost. The acceleration is correct. The force is correct. But there is a gotcha. Go back and re-read the question.

In F=ma, the F is the total force on the object from all sources. The question asks for the net force by the water on the swimmer. Can you think of something that you've missed?
 
  • #10
jbriggs444 said:
Almost. The acceleration is correct. The force is correct. But there is a gotcha. Go back and re-read the question.
In F=ma, the F is the total force on the object from all sources. The question asks for the net force by the water on the swimmer. Can you think of something that you've missed?
Thanks.

I'm not sure, is there force being exerted on the swimmer by something other than the water? Or does that calculation include force other than just what is on the swimmer? Or does it have something to do with how there is both the force the water exerts on the swimmer and that the swimmer exerts on the water?
(This is a new concept and the first homework set we have had on it so I'm not entirely certain about the concepts.)

Edit: Does this have to do with gravity? How would I take that into account?
 
Last edited by a moderator:
  • #11
Wanting to Learn said:
Does this have to do with gravity? How would I take that into account?
Yes, it has to do with gravity.
1. The net force is the sum of all the forces.
2. The net force is mass times acceleration.
3. There are two forces acting on the swimmer, gravity and water friction.
4. Put it together.
 
  • #12
So -49 is total acceleration and -9.8 is the acceleration due to gravity, so does that mean that -39.2 is the acceleration due to the water?
This would make Fwater = 65•-39.2 = -2,548N
 
  • #13
Wanting to Learn said:
So -49 is total acceleration and -9.8 is the acceleration due to gravity, so does that mean that -39.2 is the acceleration due to the water?
F=ma.
You know the acceleration -- 49 m/s^2 downward.
You know the mass -- 65 kg.
You already calculated the total force -- 3185N.

What you have not calculated is the force due to the water alone.
 
  • #14
Wanting to Learn said:
So -49 is total acceleration and -9.8 is the acceleration due to gravity, so does that mean that -39.2 is the acceleration due to the water?
This would make Fwater = 65•-39.2 = -2,548N
Say positive is "up" and negative is "down".
In what direction is the net force, up or down?
In what direction is the force of gravity, up or down?
In what direction is the force of the water, up or down?
 
  • #15
Thanks jbriggs444 for the idea to just say downward instead of trying to use negatives since all the forces I'm dealing with in this are downwards, I find that easier to think about than negatives.
 
  • #16
jbriggs444 said:
What you have not calculated is the force due to the water alone.
So would this be the way to calculate only the force from the water, without including gravity?

Ftotal = 3185N downward
Fgravity = 65•9.8 = 637N downward

Ftotal- Fgravity = Fwater
3185 - 637 = 2,548

Answer:
Fwater = 2,548N downward
 
  • #17
If all the forces are downward and the swimmer is already moving in a downward direction, the speed will increase will it not? To stop something from moving in a given direction, you need to apply a force in the opposite direction. Please reconsider my post #14.

On edit: Perhaps you are confusing velocity and acceleration.
When the acceleration and the velocity point in the same direction, the speed is increasing regardless of whether that direction is positive or negative.
When the acceleration and the velocity point in opposite directions, the speed is decreasing regardless of which is positive and which is negative.
The converse is also true.
 
  • #18
kuruman said:
If all the forces are downward and the swimmer is already moving in a downward direction, the speed will increase will it not? To stop something from moving in a given direction, you need to apply a force in the opposite direction. Please reconsider my post #14.
So is the force of the water upward?
kuruman said:
Say positive is "up" and negative is "down".
In what direction is the net force, up or down?
Is net force up? I'm confused trying to think about this because the swimmer is going down, but I think the force is pushing up because if it was pushing down then the swimmer would just keep going more down.
kuruman said:
In what direction is the force of gravity, up or down?
Gravity would be down.
kuruman said:
In what direction is the force of the water, up or down?
Reference my first replies, is the water force pushing up?

I'm not sure how to apply all this into the calculations if some things should be positive and some negative.
 
  • #19
Wanting to Learn said:
Is net force up?
Facts
1. The net force is always in the direction of the acceleration.
2. The velocity is down in this case.
3. The speed is decreasing because the swimmer eventually comes to a stop.
4. When the speed is decreasing, the acceleration is opposite to the velocity.
So ... in what direction is the net force?
Wanting to Learn said:
'm not sure how to apply all this into the calculations if some things should be positive and some negative.
The net force is the sum of all the forces. What's up is positive and what's down is negative. As you said earlier
Ftotal = Fwater + Fgravity
Just put the numbers in with the appropriate sign in front of each force and solve.
 
  • #20
kuruman said:
in what direction is the net force?
Therefore the net force is upward.
kuruman said:
What's up is positive what's down is negative.
Ftotal = + 3185
Fgravity = - 637
Fwater = + ____
Wanting to Learn said:
Ftotal- Fgravity = Fwater
Should be the same thing as
kuruman said:
Ftotal = Fwater + Fgravity
I just need to be really careful about the signs, so:
Ftotal = Fwater + Fgravity
+3185 = +Fwater + -637
+3185 +637 = +Fwater + -637 +637
+3,822 = +Fwater
So that makes final answer:
Fwater = +3,822
(upward)​
 
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  • #21
Yey! It's all very logical. :smile:
 
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  • #22
kuruman said:
Yey! It's all very logical. :smile:
Thank you!
 

FAQ: Swimmer off tower: Velocity and Force

What is the significance of studying the velocity and force of a swimmer off a tower?

The velocity and force of a swimmer off a tower are important measurements in understanding the mechanics of diving and swimming. They can help us analyze and improve the performance of swimmers and also aid in preventing injuries.

How is the velocity and force of a swimmer off a tower measured?

The velocity of a swimmer can be measured using a speedometer or by calculating the distance covered over time. The force can be measured using a force plate in the diving board or by analyzing the force exerted on the water by the swimmer's body.

What factors affect the velocity and force of a swimmer off a tower?

The velocity and force of a swimmer off a tower can be influenced by various factors such as the height of the tower, the angle of entry into the water, the speed and technique of the dive, and the body composition and strength of the swimmer.

How do velocity and force affect a swimmer's performance?

The velocity and force of a swimmer off a tower can significantly impact their performance. A higher velocity can result in a longer and more graceful dive, while a greater force can generate more power and distance in the dive. Proper control and balance of these factors can lead to a better overall performance.

How can the study of velocity and force off a tower be applied outside of swimming?

The principles of velocity and force studied in the context of a swimmer off a tower can also be applied to other sports and activities, such as high diving, cliff jumping, and even gymnastics. Additionally, understanding these concepts can also aid in designing and testing equipment for water sports and activities.

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