puzzle question for you velocity people

[Dai Capp]

A diver leaves a 5m board with an upwards vertical velocity of 2.0ms-1. How high above the water does the diver go and what velocity does he reach the water with?

Came to me from a mate who took exception to me calling his sports science degree a PE course

So, whats the answer...

Over to you...

[hornetrider]

The only real answer here is you don't know and he is better than you.

[ArtVandelay]

No horizontal velocity, wouldn't he just tt his head off the board

[Dai Capp]

hornetrider said:

The only real answer here is you don't know and he is better than you.

True enough but I can't really go back with that can I...

[jimmyjimjim]

5.204M (assuming the 5 Meter board is 5M off the surface), 2.28m/s?

Accuracy of answer to be tempered with the fact it's 6a.m. for me , and I'm not functioning well.

[BGB autosport]

Depends how heavy, surface area etc etc if you want to get picky.

As said, in answer to the question he/she would not hit the water but land back in their feet on the board.

[fathomfive]

Based on that information, he lands back on the board.

[pat_y]

Just to clarify, 2.0ms-1, this looks like 2.0ms to the power of -1 to me (0.5ms in real money).
just the velocity of upwards travel is not enough, you need to know how far upwards was travelled and i suspect you need to know mass of the subject as well. A 1 kg object travelling at 0.5ms will go higher than a 100kg object at the same speed due to gravity affecting mass/density.
Or i might be just guessing here.

[ArtVandelay]

pat_y said:

Just to clarify, 2.0ms-1, this looks like 2.0ms to the power of -1 to me (0.5ms in real money).
just the velocity of upwards travel is not enough, you need to know how far upwards was travelled and i suspect you need to know mass of the subject as well. A 1 kg object travelling at 0.5ms will go higher than a 100kg object at the same speed due to gravity affecting mass/density.
Or i might be just guessing here.

2ms-1 is the same as 2m/s (as the S is to the power of 1)

Often when calculating acceleration, I use ms-2 instead of m/s2

[Rouleur]

What if he was diving off a conveyor belt?

[Mr Gearchange]

Not enough information in the question to provide an answer.

Tell your mate he is a dick and that he should go back to wiping the sweat off exercise machines in Fitness First

[pat_y]

ArtVandelay said:

pat_y said:

Just to clarify, 2.0ms-1, this looks like 2.0ms to the power of -1 to me (0.5ms in real money).
just the velocity of upwards travel is not enough, you need to know how far upwards was travelled and i suspect you need to know mass of the subject as well. A 1 kg object travelling at 0.5ms will go higher than a 100kg object at the same speed due to gravity affecting mass/density.
Or i might be just guessing here.

2ms-1 is the same as 2m/s (as the S is to the power of 1)

Often when calculating acceleration, I use ms-2 instead of m/s2

Thank you, every day is a learning day.

[DrTre]

Greg Louganis, down, double back twister, bangs his head and in. Textbook.

[weeredmetro]

Again, assuming the 5m board is 5m above the water surface,

1)

v^2 = u^2 + 2as where v = final velocity, u = initial velocity, a = acceleration (-9.8ms^-2, gravity), s = distance

At the highest height above the board, v = 0 as the diver stops travelling up and starts travelling down.

0^2 = 2^2 + 2*(-9.8)*s
0 = 4 - 19.6s
19.6s = 4
s = 4/19.6
s = 0.204m

so diver will be 5.204 m above the water.

2) Same formula, this time we want to know v when s = -5 (ie 5m below the diving board)

v^2 = u^2 +2as
v^2 = 2^2+ 2*(-9.8)*(-5)
v^2 = 4 + 98
v^2 = 102
v = square root of 102
v = 10.0995ms^-1

These results are true provided the diver leaves the board with some horizontal velocity, otherwise as has been pointed out he will land back on the board, unless it is on a conveyor belt.

[Dai Capp]

PH has some real boffins as members, though I do like the idea of suggesting he goes back to wiping sweat off the exercise machines.

Thanks for all the responses - serious and funny - PH makes my day!

[WorAl]

weeredmetro said:

Again, assuming the 5m board is 5m above the water surface,

1)

v^2 = u^2 + 2as where v = final velocity, u = initial velocity, a = acceleration (-9.8ms^-2, gravity), s = distance

At the highest height above the board, v = 0 as the diver stops travelling up and starts travelling down.

0^2 = 2^2 + 2*(-9.8)*s
0 = 4 - 19.6s
19.6s = 4
s = 4/19.6
s = 0.204m

so diver will be 5.204 m above the water.

2) Same formula, this time we want to know v when s = -5 (ie 5m below the diving board)

v^2 = u^2 +2as
v^2 = 2^2+ 2*(-9.8)*(-5)
v^2 = 4 + 98
v^2 = 102
v = square root of 102
v = 10.0995ms^-1

These results are true provided the diver leaves the board with some horizontal velocity, otherwise as has been pointed out he will land back on the board, unless it is on a conveyor belt.

You didn't even revert to a book or use a calculator did you?

[Mr E]

ArtVandelay said:

No horizontal velocity, wouldn't he just tt his head off the board

Assuming g is 9.8ms-2, and we ignore any resistance throughout the calculation (eg, increased air resistance due to the increased speed), and we assume the diver does not clatter the board on the way back;

2ms-1 up means a drop of ~5.2m to the water under 1g of constant acceleration

About 10.1ms.


(the engineering bit of me didn't even want to remember the formula. It went along the lines of, 5m diving board, ~10ms-1 acceleration means over the first second the average speed is ~5ms. Which is almost exactly the distance the diver has to fall. If diver falls for a second at about 10ms-2 acceleration, they will be about 10ms when they hit the water.)



edit; and beaten to it.

Edited by Mr E on Monday 10th May 13:43

[TooLateForAName]

without checking the numbers weeredmetro looks to have the right approach.


The jumper starts with a kinetic energy = 1/2 mv2 (half mass x square of the velocity) and potential energy = mgh (mass x acceleration due to gravity x distance (5m)

At the top of the jump all the kinetic energy has been converted to potential energy so 1/2mv2 = mgh where h is the extra height gained due to the initial potential energy. so h = v2/2g

You then do much the same to get the terminal velocity because all the potential energy (from the 5m + the extra) is converted to kinetic energy at the water level.

[Four Cofffee]

Its a PE question, from someone with a sports science degree, so be sure to include the word 'unemployable' and the expression 'can somebody clean that st from the pool' in your answer, just so they understand you.

[the-gofer]

pat_y said:

Just to clarify, 2.0ms-1, this looks like 2.0ms to the power of -1 to me (0.5ms in real money).
just the velocity of upwards travel is not enough, you need to know how far upwards was travelled and i suspect you need to know mass of the subject as well. A 1 kg object travelling at 0.5ms will go higher than a 100kg object at the same speed due to gravity affecting mass/density.
Or i might be just guessing here.

Errr. You seem to be suggesting that you believe that mass has an effect on the speed at which an object accelerates downwards due to the Earth's gravity. Whilst this may be true at an infinitesimal level due the effect of inter-body gravitational effects it would take an extremely large mass to have an affect on the Earth itself (think, the Moon), it is not true of everyday Newtonian mechanics.

Heavier objects do not fall faster than lighter objects. As the poster above has said, the key is in GCSE level physics - v^2 = u^2 + 2as. I agree with the OP, hardly degree level "science" stuff though.