Horizontally launched projectile

[Davidllerenav]

Homework Statement

A bomb is launched from the front of a plane horizontally. The inicial velocity of the bomb is ##v##, the inicial velocity of the plane is ##u##. Find:

1. The trayectory of the bomb with respect to the ground.
2. The trayectory of the bombb with respect to the plane.
3. The trayectory of the plane with respect to the bomb.

Homework Equations

SUVAT

The Attempt at a Solution

I tried the first one like is shown in the picture. Is it right? How do I solve the others?

 

[BvU]

Hi,

Nice realistic sketch of the plane
I miss ##y_0## in your last line ? Where is ##y = 0## ?

re part 2: How about ##\ \vec r_{\rm bomb}- \vec r_{\rm plane} ## ?

 

[Davidllerenav]

Hi,

Nice realistic sketch of the plane
I miss ##y_0## in your last line ? Where is ##y = 0## ?

re part 2: How about ##\ \vec r_{\rm bomb}- \vec r_{\rm plane} ## ?

Thanks. What do you mean by ##y_0##? Isn't it equal to h?. Why ##y=0##?
For the secon one I did ##v_{b/p}=v-u##, you are doing the same but with the positions, right?

 

[BvU]

SUVAT says ##y(t) = y_0 + v_{0,y} t - {1\over 2} g t^2##. Your picture suggests ##y_0 = h##

For the second one I did

I don't know what you did, only what you asked

How do I solve the others?

 

[BvU]

could you read aloud the last line for me ?

 

[Davidllerenav]

SUVAT says ##y(t) = y_0 + v_{0,y} t - {1\over 2} g t^2##. Your picture suggests ##y_0 = h##

Yes, that's why ##y(t)=h-\frac{1}{2}gt^{2}##, since ##v_{0y}=0##

I don't know what you did, only what you asked

I said that the speed of the bomb with respect to the plane is the speed of the bomb- the speed of the plane.

View attachment 238918
could you read aloud the last line for me ?

It means that the position vector is the sum of the position on the x-axis which is ##x(t)=v_0t=ut## and the position on the y-axis which is ##y(t)=h-\frac{1}{2}gt^{2}##. Thus the trayectory would be given by ##\vec r= ut+h-\frac{1}{2}gt^{2}##.

 

[BvU]

Well well, I had difficulty reading the +h in the last line.

##y(t)=h-\frac{1}{2}gt^{2}## is good. So is ##x(t) = ut##. But I resent $$r = ut +h - g{t^2\over 2}\ .$$You can not add ##ut## and ##h-g{t^2\over 2}## as if they are numbers.

 

[Davidllerenav]

Well well, I had difficulty reading the +h in the last line.

##y(t)=h-\frac{1}{2}gt^{2}## is good. So is ##x(t) = ut##. But I resent $$r = ut +h - g{t^2\over 2}\ .$$You can not add ##ut## and ##h-g{t^2\over 2}## as if they are numbers.

I must add the ##\vec i## and ##\vec j## unit vectors right?

 

[BvU]

Yes. And then ##\vec r ## is a vector again.

 

[Davidllerenav]

Yes. And then ##\vec r ## is a vector again.

Ok. So the first one is done. Now I just need to do ##\ \vec r_{\rm bomb}- \vec r_{\rm plane}## as you said for the second part right?

 

[BvU]

I agree. You're doing fine. Bedtime for me

 

[Davidllerenav]

I agree. You're doing fine. Bedtime for me

Ok, thanks! One last question, for the third part, I just need to do the same thing from the second part but it would be ##\ \vec r_{\rm plane}- \vec r_{\rm bomb}##, right?

 

[Davidllerenav]

I agree. You're doing fine. Bedtime for me

Can this problem be solved using Galilean transformations?

 

[BvU]

Yes

 

[Davidllerenav]

Yes

And how would it be if we solve it with Galilean transformations?

 

[Davidllerenav]

Yes

I ended up with ##\vec r = (V-U)t \vec i -(g\frac{t^{2}}{2})\vec j## in the case of the bomb with respect to the plane, and ##\vec r = (U-V)t \vec i +(g\frac{t^{2}}{2})\vec j## in the case of the plane with respect to the bomb, am I right?

 

[BvU]

One way to find out: hand it in !