How Do Two Divers' Speeds Compare When Jumping from a 3m Platform?

Click For Summary

Homework Help Overview

The problem involves two divers jumping from a 3-meter platform, with one diver jumping upward at a speed of 1.8 m/s and the other stepping off the platform as the first diver descends. The questions focus on determining their speeds upon hitting the water and which diver reaches the water first.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss calculations for the final speeds of both divers as they hit the water, using kinematic equations. There are attempts to clarify the sign conventions used in the calculations and the values of gravitational acceleration. Some participants suggest checking for rounding errors and the assumptions made regarding the timing of the divers' jumps.

Discussion Status

Participants are actively engaging with the problem, sharing their calculations and questioning the assumptions made in the original attempts. There is recognition of potential discrepancies in the expected answers, and some guidance is offered regarding the use of consistent sign conventions and the correct interpretation of the problem setup.

Contextual Notes

There is mention of a possible difference in the value of gravitational acceleration used in calculations, as well as the need to clarify the definition of speed versus velocity in the context of the problem.

negation
Messages
817
Reaction score
0

Homework Statement



Two divers jumped from a 3m platform. One jumps upward at 1.8ms^-1 and the second steps off the platform as the first passes it on the way down.

a) What are their speeds as they hit the water?
b) Which hits the water first and by how much?

Homework Equations





The Attempt at a Solution




a) What are their speeds as they hit the water?

Diver 1:
Calculate distance diver 1 travels as he jumps 1.8ms^-1 upwards off the platform.

vf^2 - vi^2 = 2g(yf - yi)
0 - (18ms^-1)^2 = 2(-9.8ms^-2)(yf - yi)
yf - yi = 0.165m
3 m + 0.165m = 3.165m
vf^2 - vi^2 = 2g(yf - yi)
vf = 7.8ms^-1

diver 1 hits water at 7.8ms^-1

Diver 2:
vf^2 - vi^2 = 2g(yf - yi)
vf^2 - 0 = 2(-9.8ms^-2)(-3m)
vf = 7.6ms^-1

Diver 2 hits water at 7.6ms^-1


b) Which hits the water first and by how much?

Diver 1:

At the instant at which diver 1 is 3.165m off the ground, vi = 0
-3.165m = 0.5(0ms^-1 + (-7.8ms^-1))t
t = 0.8s

Diver 2:

-3 = 0.5(0ms^-1 + (-7.6ms^-1))t
t = 0.78s

My answers are wrong according to the book:
a) -7.67ms^-1
b) 0.162s
 
Last edited:
Physics news on Phys.org
negation said:

Homework Statement



Two divers jumped from a 3m platform. One jumps upward at 1.8ms^-1 and the second steps off the platform as the first passes it on the way down.

a) What are their speeds as they hit the water?
b) Which hits the water first and by how much?

Homework Equations





The Attempt at a Solution




a) What are their speeds as they hit the water?

Diver 1:
Calculate distance diver 1 travels as he jumps 1.8ms^-1 upwards off the platform.

vf^2 - vi^2 = 2g(yf - yi)
0 - (18ms^-1)^2 = 2(-9.8ms^-2)(yf - yi)
yf - yi = 0.165m
3 m + 0.165m = 3.165m
vf^2 - vi^2 = 2g(yf - yi)
vf = 7.8ms^-1

diver 1 hits water at 7.8ms^-1

Diver 2:
vf^2 - vi^2 = 2g(yf - yi)
vf^2 - 0 = 2(-9.8ms^-2)(-3m)
vf = 7.6ms^-1

Diver 2 hits water at 7.6ms^-1


b) Which hits the water first and by how much?

Diver 1:

At the instant at which diver 1 is 3.165m off the ground, vi = 0
-3.165m = 0.5(0ms^-1 + (-7.8ms^-1))t
t = 0.8s

Diver 2:

-3 = 0.5(0ms^-1 + (-7.6ms^-1))t
t = 0.78s

My answers are wrong according to the book:
a) -7.67ms^-1
b) 0.162s

For the first part, I would check what value of g the textbook expects you to use, and whether you had round-off errors in intermediate steps. g = -9.81m/s^2 gives the expected answer to 3 s.f.

Your working in the first part looks OK, but I would be careful of your sign convention. Define up to be positive, and down to be negative. That may explain why the answer key has a negative sign (but they mentioned speed, which shouldn't have a sign - if they'd said velocity, they'd be correct).

Note that you don't actually have to calculate the point at which Diver 1 comes to momentary rest. Just use initial velocity = +1.8m/s and g = -9.81m/s^2 to get the final velocity in one step. Remember that the s term (or y term, as you're using it) is actually displacement and not total distance. Minimise the number of steps you take to get your answers, and there will be less chance of error.

As for the second part, I'm assuming you're using ##s = \frac{(u+v)t}{2}## (state the equation you're using - always!).

For diver 1, you seem to be working out the time taken to reach the water *from the time he is at maximum height and is momentarily at rest*. Is this really the time you want for diver 1? Remember that diver 2 steps off the platform only as diver 1 passes him.
 
  • Like
Likes   Reactions: 1 person
Curious3141 said:
For the first part, I would check what value of g the textbook expects you to use, and whether you had round-off errors in intermediate steps. g = -9.81m/s^2 gives the expected answer to 3 s.f.

Your working in the first part looks OK, but I would be careful of your sign convention. Define up to be positive, and down to be negative. That may explain why the answer key has a negative sign (but they mentioned speed, which shouldn't have a sign - if they'd said velocity, they'd be correct).

I'm still getting -7.88ms^-1 after rounding off to 3SF
If the answer states -7.67ms^-1 then something is wrong
 
negation said:
I'm still getting -7.88ms^-1 after rounding off to 3SF
If the answer states -7.67ms^-1 then something is wrong

The question asks for speeds, plural. Diver 1's speed is 7.88m/s downward (so velocity = -7.88m/s). I'm getting that too, so I think you're right.

Diver 2's speed is 7.67m/s downward (v = -7.67m/s). Are you getting this?
 
Curious3141 said:
The question asks for speeds, plural. Diver 1's speed is 7.88m/s downward (so velocity = -7.88m/s). I'm getting that too, so I think you're right.

Diver 2's speed is 7.67m/s downward (v = -7.67m/s). Are you getting this?

I am.

I'll work on part (b) now
 

Similar threads

Replies
7
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
9
Views
3K
Replies
5
Views
2K
Replies
6
Views
3K
Replies
5
Views
3K
  • · Replies 24 ·
Replies
24
Views
4K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K