How Does Drag Affect Projectile Speed Over Time?

[craig.16]

Homework Statement

A projectile of mass 0.1 kg traveling horizontally through the air with speed v(t) experiences
a drag force DT (t) whose magnitude is given by:
D (t ) 0.001 [v (t )] 2
At one instant, the projectile’s speed is measured to be 100 m/s. Calculate the speed of the
projectile after a time of 3 s has elapsed, during which the projectile continues to travel
horizontally.

Homework Equations

D(T)=1/2*C*(rho)*A*(v^2) or D(T)=b(v^2)

The Attempt at a Solution

I tried inputting the the speed into the given equation and got D(T)= 10. Other than that I don't know what else to do. I used the equation:
v=sqrt(mg/b)*(tanh((t)sqrt(bg/m)) which gave me 23m/s but that seems rather unlikely as it couldn't have dropped 77m/s in 3 seconds. Also this equation is for vertical not horizontal so it is most likely wrong anyway but I don't know how to properly approach this question. Any help will be much appreciated.

 

[Beaker87]

Homework Statement

A projectile of mass 0.1 kg traveling horizontally through the air with speed v(t) experiences
a drag force DT (t) whose magnitude is given by:
D (t ) 0.001 [v (t )] 2
At one instant, the projectile’s speed is measured to be 100 m/s. Calculate the speed of the
projectile after a time of 3 s has elapsed, during which the projectile continues to travel
horizontally.

The only force acting on the projectile is the drag force. Try to use Newton's second Law:
$$F=ma$$
or, in terms of velocity:
$$F=m \frac{dv}{dt}$$

Then you can divide the force by the mass, multiply by dt, divide by v² and integrate both sides. Make sure you make the drag force negative.

 

[craig.16]

That seems to make a lot more sense thanks. However its the dividing by v^2 that throws me off a bit. Am I taking D(T) as F but negative because if so I used my value of (-10) and got an even smaller answer but only due to the division of v^2 as I don't know specifically where that gets divided from. Otherwise I would get a massive decrease. To make things sound less confusing I did the following:

F=D(T)

-10=0.1(dv/dt)
-100dt=dv
integrating both sides I get:
-100t=v
v=-100(3)=-300m/s

At what point did I go wrong?

 

[Beaker87]

That seems to make a lot more sense thanks. However its the dividing by v^2 that throws me off a bit. Am I taking D(T) as F but negative because if so I used my value of (-10) and got an even smaller answer but only due to the division of v^2 as I don't know specifically where that gets divided from. Otherwise I would get a massive decrease. To make things sound less confusing I did the following:

F=D(T)

-10=0.1(dv/dt)
-100dt=dv
integrating both sides I get:
-100t=v
v=-100(3)=-300m/s

At what point did I go wrong?

Your mistake was D(t)=-10

D is a function of t, and changes as time goes on.

Think of it this way:
As the drag force is applied, it slows the object down; as the object slows, the drag force decreases.

Therefore when you substitute it into the equation, F=ma, you need to substitute it in as the function, -D(t)=-0.001v(t)²
(I made it negative because the force is going to be negative)

This gives you the equation:
$$-0.001v(t)^{2}=m \frac{dv}{dt}$$

Once you solve for v(t), you can then use your initial value of v(0)=100.

 

[craig.16]

Am I right by saying that it gets rearanged and integrated to:
-0.01v(t)^2t=v(0)
If so its the minus sign that ruins it when it comes to square rooting the answer unless I have to take into account imaginary numbers which I am pretty sure I don't need to. I tried with it being positive and I got 57.7m/s which is more realistic as compared to my previous attempt. Also note that I have divided the mass already hence the -0.01 instead of -0.001. Finally if I have done something stupid am I at least right in thinking that v(t)^2 is not the same as v*(t)^2 as v is a function of t like f(x) and y. Apologies if this feels like pretty trivial stuff.

 

[Beaker87]

Am I right by saying that it gets rearanged and integrated to:
-0.01v(t)^2t=v(0)
If so its the minus sign that ruins it when it comes to square rooting the answer unless I have to take into account imaginary numbers which I am pretty sure I don't need to. I tried with it being positive and I got 57.7m/s which is more realistic as compared to my previous attempt. Also note that I have divided the mass already hence the -0.01 instead of -0.001. Finally if I have done something stupid am I at least right in thinking that v(t)^2 is not the same as v*(t)^2 as v is a function of t like f(x) and y. Apologies if this feels like pretty trivial stuff.

Something went wrong with your rearranging and integrating.

When you rearrange the function, you get:
$$-0.001dt=m\frac{dv}{v^2}$$

Now integrate both sides, and add an arbitrary constant, C, to either side (it does not matter which side)

Now solve for v(t). (Get v alone on one side of the equation)

Next, using the fact that v(0)=100, solve for C.

Finally, find v(3).


You should find an answer of:

Spoiler

25m/s, so don't be alarmed when the projectile lost 3/4 of its speed in 3 seconds, the drag coefficient used here was quite high.

 

[craig.16]

Yeah I realize that I stupidly forgot to carry over the v^2. after integrating properly this time around I got the answer. Thanks for taking the time to help me with my question.