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

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Two divers jumped from a 3m platform, with one jumping upward at 1.8 m/s and the other stepping off as the first descends. Diver 1 hits the water at a speed of approximately 7.88 m/s, while Diver 2 reaches a speed of about 7.67 m/s. Diver 1 takes longer to reach the water, hitting it after approximately 0.8 seconds, compared to Diver 2's 0.78 seconds. The calculations emphasize the importance of using the correct gravitational constant and maintaining consistent sign conventions. The discussion highlights the need for careful attention to detail in physics problems to avoid errors in speed and timing calculations.
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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
 
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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.
 
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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
 
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