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

• negation
In summary: I am.I'll work on part (b) now.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.I am using the equations = ut + 1/2at^2For diver 1, I am calculating the time it takes for him to reach the water after he passes diver
negation

## 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?

## 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:
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?

## 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.

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

## What is Diver Speed Kinematics motion?

Diver Speed Kinematics motion is a branch of physics that studies the motion of divers, specifically focusing on their speed and how it changes over time.

## What factors affect a diver's speed?

There are several factors that can affect a diver's speed, including their body position, the force of gravity, and the resistance of the water.

## How is Diver Speed Kinematics motion calculated?

Diver Speed Kinematics motion is calculated using mathematical equations that take into account the initial speed of the diver, the time interval, and any forces acting on the diver.

## What is the difference between average speed and instantaneous speed in Diver Speed Kinematics motion?

Average speed in Diver Speed Kinematics motion is the total distance traveled divided by the total time, while instantaneous speed is the speed of the diver at a specific moment in time.

## How is Diver Speed Kinematics motion used in practical applications?

Diver Speed Kinematics motion is used in practical applications such as analyzing the performance of divers in competitions, designing more efficient diving techniques, and understanding the physics of diving accidents. It can also be applied to other sports and activities involving motion in fluids.

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