# How Is Delta_h Calculated in Physics Problems?

• annamal
In summary, the answer to part b is that the diamond falls 1.65 meters before it reaches 90% of its terminal speed.
annamal
Homework Statement
A small diamond of mass 10.0 g drops from a swimmer’s earring and falls through the water, reaching a terminal velocity of 2.0 m/s. a) Assuming the frictional force on the diamond obeys what is b? (b) How far does the diamond fall before it reaches 90 percent of its terminal speed?
Relevant Equations
v_t = terminal velocity
a) -m*g + b*v = m*a = 0 for terminal velocity
b = m*g/v_t
b) My question is here:
0.9v_t = v0 + a*t = a*t
t = 0.9v_t/a
delta_h = v0*t + 0.5*a*t^2 = 0.5*a*t^2 = 0.5*((0.9*v_t)^2)/a =
where a = (-m*g + b*v)/m
delta_h = (0.5*((0.9*v_t)^2)*m)/(-m*g + b*v)
plugging in everything delta_h = 1.65 m which is not what the answer is. Please help
b) My question is here!
0.9v_t = v0 + a*t = a*t
t = 0.9v_t/a
delta_h = v0*t + 0.5*a*t^2 = 0.5*a*t^2 = 0.5*((0.9*v_t)^2)/a =
where a = (-m*g + b*v)/m
delta_h = (0.5*((0.9*v_t)^2)*m)/(-m*g + b*v)
plugging in everything delta_h = 1.65 m which is not what the answer is. Please help

Could you clarify this part?
"...Assuming the frictional force on the diamond obeys what is b? (b) How far..."

annamal said:
0.9v_t = v0 + a*t = a*t
This is true if acceleration is constant. Is acceleration constant?

annamal said:
A small diamond of mass 10.0 g
That's a 50 carat diamond! It'd be as big as an acorn!

Delta2, hutchphd, Bystander and 1 other person
DaveC426913 said:
That's a 50 carat diamond! It'd be as big as an acorn!
Good observation. This also brings to mind another point though, the total force is not mg-bv ...

As an aside, a 10-g diamond is not exactly "small". Considering that there are 5 carats to a gram, a 10-gram diamond would be 50 carats.

"A 50-carat white diamond has fetched 6.5 million Swiss francs (dollars) at a Christie's auction of jewelry, with the auctioneer saying an anonymous trader snapped up the rare stone."
Source: https://www.dailysabah.com/life/2018/05/17/50-carat-diamond-sold-for-65m-at-christies-auction

How stupid can one be to go swimming with of these hanging from one's ear? I believe that physics problems ought to be reasonably realistic.

On edit: I see @DaveC426913 preempted me.

Delta2
Orodruin said:
This is true if acceleration is constant. Is acceleration constant?
a = (-m*g + b*v)/m --> acceleration is dependent on velocity. why does acceleration have to be constant?

Orodruin said:
Good observation. This also brings to mind another point though, the total force is not mg-bv ...
No, it is. The solution said so.

annamal said:
Homework Statement:: A small diamond of mass 10.0 g drops from a swimmer’s earring and falls through the water, reaching a terminal velocity of 2.0 m/s. a) Assuming the frictional force on the diamond obeys what is b? (b) How far does the diamond fall before it reaches 90 percent of its terminal speed?
Relevant Equations:: v_t = terminal velocity
a) -m*g + b*v = m*a = 0 for terminal velocity
b = m*g/v_t
b) My question is here:
0.9v_t = v0 + a*t = a*t
t = 0.9v_t/a
delta_h = v0*t + 0.5*a*t^2 = 0.5*a*t^2 = 0.5*((0.9*v_t)^2)/a =
where a = (-m*g + b*v)/m
delta_h = (0.5*((0.9*v_t)^2)*m)/(-m*g + b*v)
plugging in everything delta_h = 1.65 m which is not what the answer is. Please help

b) My question is here!
0.9v_t = v0 + a*t = a*t
t = 0.9v_t/a
delta_h = v0*t + 0.5*a*t^2 = 0.5*a*t^2 = 0.5*((0.9*v_t)^2)/a =
where a = (-m*g + b*v)/m
delta_h = (0.5*((0.9*v_t)^2)*m)/(-m*g + b*v)
plugging in everything delta_h = 1.65 m which is not what the answer is. Please help
Assuming the frictional force on the diamond obeys -bv what is b?

Lnewqban said:
Could you clarify this part?
"...Assuming the frictional force on the diamond obeys what is b? (b) How far..."
Assuming the frictional force on the diamond obeys -bv what is b?

Lnewqban
annamal said:
No, it is. The solution said so.
That it is in the solution does not make it correct. In this case, the solution is wrong. It would be correct if the water density was very small compared to the diamond density, but that is not the case. The specific gravity of diamond is ca 3.5, which is not that much larger than 1.
annamal said:
a = (-m*g + b*v)/m --> acceleration is dependent on velocity. why does acceleration have to be constant?
The formula you quoted:
$$v = v_0 + at$$
assumes that ##a## is constant. It is not valid if ##a## is not constant.

Orodruin said:
That it is in the solution does not make it correct. In this case, the solution is wrong. It would be correct if the water density was very small compared to the diamond density, but that is not the case. The specific gravity of diamond is ca 3.5, which is not that much larger than 1.

The formula you quoted:
$$v = v_0 + at$$
assumes that ##a## is constant. It is not valid if ##a## is not constant.
Ok, I found the formula v = v0*e^(-bt/m) but I am confused because isn't v0 the initial velocity 0?

annamal said:
Ok, I found the formula v = v0*e^(-bt/m) but I am confused because isn't v0 the initial velocity 0?
That is the formula without gravitation or buoyancy. It is not applicable here. You can easily see this from the fact that it goes to zero (and not the terminal velocity) as time goes to infinity.

Orodruin said:
That is the formula without gravitation or buoyancy. It is not applicable here. You can easily see this from the fact that it goes to zero (and not the terminal velocity) as time goes to infinity.
Late, but to solve part b, I did, ##0.9v = \int_0^t(mg-bv)dt## where b= 0.049, but I don't know how to solve that equation

annamal said:
Late, but to solve part b, I did, ##0.9v = \int_0^t(mg-bv)dt## where b= 0.049, but I don't know how to solve that equation
You can't solve it in that form because of the unknown function v(t) inside the integral. You need to get all the occurrences of v together.
Taking it back the the differentIal form, ##0.9\dot v = mg-bv##. Now get both v's on the same side. There are two ways to do that. Both work, but the way forward may be more obvious with one than with the other.

Last edited:
haruspex said:
You can't solve it in that form because of the unknown function v(t) inside the integral. You need to get all the occurrences of v together.
Taking it back the the differentIal form, ##0.9\dot v = mg-bv##. Now get both v's on the same side. There are two ways to do that. Both work, but the way forward may be more obvious with one than with the other.
ok, I am rusty with my diff eq, I get ##0.9\frac{dv}{dt} + bv = mg##... and then what?

Then that is a linear ODE with constant coefficients.

annamal said:
ok, I am rusty with my diff eq, I get ##0.9\frac{dv}{dt} + bv = mg##... and then what?
Of course, you have picked the harder one. Can you see the other way to get all the v's on the same side?

haruspex said:
Of course, you have picked the harder one. Can you see the other way to get all the v's on the same side?
Oh nvm. I found the derivation in the book and got the correct answer. Thanks anyways.

annamal said:
Oh nvm. I found the derivation in the book and got the correct answer. Thanks anyways.
Fwiw, the other way is to divide both sides by mg-bv. Then both sides are integrable.

Last edited:
Delta2
annamal said:
Oh nvm. I found the derivation in the book and got the correct answer. Thanks anyways.
Getting the right answer is not the end goal. The end goal is understanding how you got it and why it is the right answer. Just looking up the solution is generally not a good approach.

nasu

## 1. What is Delta_h in math?

Delta_h is a mathematical symbol that represents the change in height or altitude of a given object or function.

## 2. How do you solve for Delta_h?

To solve for Delta_h, you need to have an initial value and a final value for the height or altitude. Then, you subtract the initial value from the final value to find the difference or change in height.

## 3. What is the formula for solving for Delta_h?

The formula for solving for Delta_h is: Delta_h = final value - initial value.

## 4. Can Delta_h be negative?

Yes, Delta_h can be negative. This indicates a decrease or downward change in height or altitude.

## 5. What are some real-life applications of solving for Delta_h?

Solving for Delta_h can be useful in many fields, such as engineering, physics, and geography. For example, it can be used to calculate the change in elevation of a roller coaster or the change in altitude of an airplane during a flight.

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