Diving airplane as a projectile

[rudransh verma]

Homework Statement

A certain airplane has a speed of 290km/h and is diving at an angle of $\theta=30 degrees$ below the horizontal when the pilot releases the radar decoy. The horizontal distance between the release point and the hitting point is $d=700m=0.7 km$. A) how long is the decoy in the air B) how high was the release point?

Relevant Equations

$s=ut+\frac12at^2$

The first part is simple. $10 secs = 0.0027 hrs$. $$0.7=290\cos30t$$
For the second part $$s=-290\sin30t-\frac129.8(0.0027)^2$$
But I am getting wrong result.

 

[haruspex]

below the horizontal

So what is the initial vertical velocity?

 

[rudransh verma]

So what is the initial vertical velocity?

$-290\sin30$

 

[haruspex]

$-290\sin30$

You also have a problem with units. Try filling in the units in your equations.

 

[rudransh verma]

You also have a problem with units. Try filling in the units in your equations.

The answer is wrong. It’s not just a sign.
Am I to take - of angles because it’s clockwise?

 

[BvU]

Do you know what units are ?
E.g. what are the units in your $$s=ut+\frac12at^2\quad ?$$

 

[rudransh verma]

I

Do you know what units are ?
E.g. what are the units in your $$s=ut+\frac12at^2\quad ?$$

I gotch ya. It’s g.

 

[BvU]

Do me a favour, answer the question

 

[rudransh verma]

Do me a favour, answer the question

$-391.6 m$ It does not match with the given answer.

 

[jbriggs444]

You've not answered the question:

Do you know what units are ?
E.g. what are the units in your $$s=ut+\frac12at^2\quad ?$$

What are the units on $s$?
What are the units on $u$?
What are the units on $t$?
What are the units on $a$?

$-391.6 m$ It does not match with the given answer.

Spitting out a wrong answer without showing us how you got that answer does not help very much. It is difficult to spot the errors in your work when you do not show us your work.

 

[Steve4Physics]

Hi @rudransh verma. @BvU asked

Do you know what units are ?
E.g. what are the units in your $$s=ut+\frac12at^2\quad ?$$

to which your first answer was 'g' and your second answer was $-391.6m$.

I'm not being rude but do you see the problem.? It's important (and courteous) to answer the question asked to the best your ability even if, sometimes, the answer is 'I don't know'.

But supplying the answer to a different question is a real hindrance. You're not a politician are you?

 

[rudransh verma]

What are the units on s?
What are the units on u?
What are the units on t?
What are the units on a?

$km\frac{km}{h} h \frac{km}{h^2}$

 

[jbriggs444]

$km,\frac{km}{h},h,\frac{km}{h}^2$

So g is 9.8 km/h²?

Putting the unit names in $LaTeX$ has the effect of putting them in italics and removing white space. Probably not the best choice for presenting just the units alone.

Using \text{km}, \frac{\text{km}}{\text{h}}, etc could succeed. That switches off italics.
Also, you can put a backslash in front of blank if you want to have white space.

Edit: The units for acceleration were wrong as written. You want $\frac{\text{km}}{\text{h}^2}$, not $\frac{km}{h}^2 = \frac{{km}^2}{h^2}$

 

[rudransh verma]

So g is 9.8 km/h²?

The backslash goes in front, \frac, not frac\.

127,008 km/h^2

 

[jbriggs444]

Great, so now can you show us your work that leads to the altitude of 391.6 meters below ground for the release point? [I notice that this result was reported in meters, not kilometers, what's up with that?]

 

[Steve4Physics]

Hi again @rudransh verma. Just for information. in Post #1 you wrote:
"10 secs = 0.0027 hrs"

a) It is wrong. 10 secs ≈ 0.0028 hrs (if rounded to 2 significant figures which, anyway, is not a good idea if the value is to be used in a calculation).

b) I don't see why you stated this!
[Edit. Ah, I get it. You calculated the time of flight as 0.002787hrs, incorrectly stated it as 0.0027hrs and converted it to seconds, giving 10s, Then wrote some of the working after the answer.]

Also, to get the degree symbol in Latex, note that \circ produces a small circle, so ^\circ produces $^\circ$.

 

[jbriggs444]

$-391.6 m$ It does not match with the given answer.

Reverse engineering, it seems that this answer was obtained by using the incorrect value for g: 9.8 km/h².

That gives -0.3915 km (-391.5 meters) due to the initial downward velocity over the course of the [improperly rounded] 0.0027 hours.

The additional downward displacement due to gravity is negligible due to the error in units.

Then we have the sign error due to reporting the downward displacement rather than the upward launch altitude.

 

[haruspex]

Perhaps you do not understand what is meant by showing the units in the equations.

You start with

$s=ut+\frac 12at^2$

This does not need units because the dimensions are part of the variables.
But, correspondingly, when numbers replace some of them the units come too:

For the second part $$s=-290\sin30t-\frac129.8(0.0027)^2$$

That should read $$s=-(290km/h)\sin30t-\frac129.8(m/s^2)(0.0028h)^2$$
$=-290(km/h)(1000m/km)/(3600s/h)\frac 12(10s)-\frac129.8(m/s^2)(0.0028h)^2(3600s/h)^2$
$=-290/3.6m/s(5s)-4.9(m/s^2)(10s)^2$
Etc.
In this way, you can be much surer of making all appropriate conversions and of making them correctly.

The other mistake you might still be making is what s represents. In your equations, it is the vertical displacement of the decoy, up measured as positive. That is not exactly what the question asks for.

 

[rudransh verma]

Great, so now can you show us your work that leads to the altitude of 391.6 meters below ground for the release point? [I notice that this result was reported in meters, not kilometers, what's up with that?]

$s=-290\sin30(0.0028)-\frac12(127008)(0.0028)^2$
$$s=-0.406-0.4978$$
$$ -0.9038 kilometers = -903.8 meters $$
Taking $t= 0.00278 h$ I get $s = -893.8 meters$

Since it’s asking height so neglecting - sign $$s = 893.8 meters $$
I got it very close. Thanks

I have a doubt. By any chance should we have taken $\theta$ -ve since it’s clockwise.

 

[jbriggs444]

I have a doubt. By any chance should we have taken $\theta$ [negative] since it’s clockwise.

You could do things that way.

When you did it your way, you put in the minus sign by hand, presumably because you'd already noticed that "the angle is below the horizontal". That is perfectly fine.

You could equally well have let the minus sign come in because $\sin -30^\circ$ is negative.

 

[rudransh verma]

You could do things that way.

When you did it your way, you put in the minus sign by hand, presumably because you'd already noticed that "the angle is below the horizontal". That is perfectly fine.

You could equally well have let the minus sign come in because $\sin -30^\circ$ is negative.

I took -initial velocity in y direction and + $\theta$. My doubt is shouldn’t we take $\sin(-30)$ (because it’s clockwise) and -initial velocity(because body is moving in 4 quadrant )which would make it positive?
$-290\sin(-30)$

 

[jbriggs444]

I took -initial velocity and +$\theta$.

I am not sure that I understand what you are getting at.

It seems that you are questioning how you arrived at the equation:$$s=-290\sin30(0.0028)-\frac12(127008)(0.0028)^2$$and wondering whether you should have gotten it some other way.

Surely you started with the SUVAT equation: $s=ut + at^2$

So you are questioning the way you determined the initial velocity $u$. Should you have written it as $-290 \frac{\text{km}}{\text{h}} \sin 30^\circ$ or as $290 \frac{\text{km}}{\text{h}} \sin -30^\circ$

Either way, you are extracting the vertical component of a vector velocity. Either way you are adopting a coordinate system in which up is positive.

In the one approach, you are seeing that the angle is below the horizontal, so you compute the vertical component by multiplying by the negative of the sine of a positive angle.

In the other approach, you are seeing that the angle is in the fourth [or third if the plane is flying left] quadrant, and you multiply by the sine which you know will turn out to be negative.

You get the same number either way. It does not matter which approach you use. Both are correct.

 

[rudransh verma]

So you are questioning the way you determined the initial velocity u. Should you have written it as −290kmhsin⁡30∘ or as 290kmhsin−30∘

I am asking shouldn't it be -290sin(-30) because the velocity should be negative(that why -290) and its below horizontal angle(thats why -30).

 

[Steve4Physics]

I am asking shouldn't it be -290sin(-30) because the velocity should be negative(that why -290) and its below horizontal angle(thats why -30).

But -290sin(-30º) is positive! Your vertical component would be in the wrong direction.

By the way, you might have found it easier to convert 290km/hr to m/s at the start, and then done the rest of the problem in metres and seconds.

 

[rudransh verma]

But -290sin(-30º) is positive! Your vertical component would be in the wrong direction.

ya that's the problem. So I should make it $290\sin(-30)$ which will become -290sin30. Now it looks correct in the -y axis.
I want to clear one thing its the angle or the sign(I guess sign) which sets the direction of the components and this is polar coordinate system.
In that sense we should have taken 290cos(-30)= 290cos30 in the first part. Its in positive x axis.

 

[jbriggs444]

I am asking shouldn't it be -290sin(-30) because the velocity should be negative(that why -290) and its below horizontal angle(thats why -30).

Absolutely not. That would be double dipping. You'd get the absurd result that the vertical component of velocity for a diving plane would be positive (negative times negative is positive).

If you want to do things in a more technically correct way, you can compute the vertical component of a vector by taking the dot product of the vector with a vertical direction vector.

You can evaluate the dot product as the magnitude of the one vector times the magnitude of the other vector times the cosine of the angle between them: $$\vec{a} \cdot \vec{b} = |a| |b| \cos \theta$$

The magnitude of a vector is always positive. The magnitude of a velocity vector is called "speed". We are given that the speed is 290 km/h in this case.

The magnitude of a direction vector is always one, by definition.

So in the case at hand, the dot product we are evaluating becomes:$$u_y = |\vec{u}| |\hat{j}| \cos \theta = |\vec{u}| \cos \theta = 290 \frac{\text{km}}{\text{h}} \cos 120^\circ$$where $\hat{j}$ is a vertical direction vector and $\theta$ is now an angle measured from the vertical.

Normally we skip this level of intense formality and just say "30 degrees below the horizontal -- multiply by minus 1/2 and be done with it".

 

[jbriggs444]

I want to clear one thing its the angle or the sign(I guess sign) which sets the direction of the components and this is polar coordinate system.

You are free to choose any coordinate system that you like.

In this case, you were given information in terms of a magnitude and direction (speed and angle below the horizontal). The very first thing you set about doing was converting that magnitude and direction to a horizontal component of velocity.

That means that you were working in a cartesian coordinate system and converting your inputs to fit.

 

[BvU]

At the risk of sounding like a broken record:

It would have been very useful to accompany the work with a sketch that shows the situation and makes clear what coordinate system was chosen ​

​

In addition

When there is no good reason NOT to work in SI units, work in SI units ​

​

$\$​

 

[rudransh verma]

You are free to choose any coordinate system that you like.

In this case, you were given information in terms of a magnitude and direction (speed and angle below the horizontal). The very first thing you set about doing was converting that magnitude and direction to a horizontal component of velocity.

That means that you were working in a cartesian coordinate system and converting your inputs to fit.

That means I converted the polar coordinates to cartesian coordinates. From $(290, -30)$ to $(290\cos(-30), 290\sin(-30))$

 

[rudransh verma]

In this case, you were given information in terms of a magnitude and direction (speed and angle below the horizontal). The very first thing you set about doing was converting that magnitude and direction to a horizontal component of velocity.

I have a doubt. When finding resultant vectors from a set of given vectors then what I have learned is take the triangles one by one and find the components by taking theta positive and after computing the magnitude of components put signs according to which direction they lie. After that when you find all the $i cap/hat$ and $j cap/hat$ add them and get the resultant.

Can I find it like what I did here in this thread by taking signs with the theta angle. Will I get the same components with proper signs.

What is the correct technical way to do it?

 

[jbriggs444]

I have a doubt. When finding resultant vectors from a set of given vectors then what I have learned is take the triangles one by one and find the components by taking theta positive and after computing the magnitude of components put signs according to which direction they lie. After that when you find all the $\hat i$ and $\hat j$ add them and get the resultant.
Can I find it like what I did here in this thread by taking signs of the theta angle. Will I get the same components with proper signs.
What is the correct technical way to do it?

If you want a "hat" on a unit vector, you use \hat, e.g. \hat i which renders as $\hat i$

Yes, you can add a list of vectors by converting them all to components, adding the x and y components separately and then reporting the result either as $(\sum x, \sum y)$ or as $\hat i \sum x + \hat j \sum y$. [You'd use actual numbers and not sigmas, of course]

Yes, you can find the components of a vector in any manner that makes sense to you.

One method is:

"I know the component is negative since the angle is below the horizontal. I know I want the sine since I am looking for the length of the side opposite my angle. So I multiply the [positive] length by the [positive] sine and then negate because I already figured out that I want a negative result".

 

[rudransh verma]

So I multiply the [positive] length by the [positive] sine and then negate because I already figured out that I want a negative result".

How about doing it in a more textbook sense. By taking $\sin of -\theta$ if it is below the horizontal. Will I get the same result like -vertical component?

 

[jbriggs444]

How about doing it in a more textbook sense. By taking $\sin of -\theta$ if it is below the horizontal. Will I get the same result like -vertical component?

Yes. $-(290 \sin 30 ^\circ) = 290 \sin -30 ^\circ$

 

[rudransh verma]

Yes. $-(290 \sin \theta) = 290 \sin -\theta$

but what about if my vector is in second or third quadrant?

I know we usually take + magnitude multiple with positive sin/cos of angle and then put signs as - and + to x and y components in second quadrant because we know that’s how they exist in that quadrant in Cartesian coordinate system.

But how about not following this method and somehow take something like magnitude multiplied with sin of original angle measured from +x axis?

 

[jbriggs444]

but what about if my vector is in second or third quadrant?

So your vector is in the second quadrant. You can draw your triangle and measure the angle as 30 degrees.

You eyeball the situation and decide that the result will be positive. Using the triangle method, you find $+(290 \sin 30 ^\circ)$. Using your polar coordinates, you find $290 \sin 150 ^\circ$. Same result either way.

Or you are in the third quadrant. You draw your triangle and measure the angle as 30 degrees. You eyeball the situation and decide that the result will be negative. Using the triangle method you find $-(290 \sin 30 ^\circ)$. Using polar coordinates, you find $290 \sin 210 ^\circ$. Same result either way.