Tricky conceptual Projectile motion question

[Rubberduck2005]

Homework Statement

Two robbers one on the ground one on roof, point their guns at each other at fire simultaneously. Show that their bullets will collide unless the calculated collision point is underground, which is not allowed. The guy on the ground fires with a velocity v1 and the other at a velocity v2, show that the condition for colliding is v1+v2 > gd/v1sin(2Q) where g is 9.81ms^-2 and d is the horizontal between the two and Q is the angle of elevation of the line joining the two robbers.

Relevant Equations

g=9.81ms^-2 and Y(t) and X(t) is the motion of the robber on the roof whilst y(t) and x(t) are the robbers on the ground

So far all I have determined is the equations of motion for the two and that is as follows. It is trivial that y(t)=v1sin(Q)t -gt^2/2 and that x(t)=v2cos(Q)t. Now the angle that is anticlockwise from the negative horizontal of the robber is 90 - Q using basic trigonometry, using this we can determine that the velocity projected onto the plane will be the negative magnitude of v2cos(90 - Q)=v2sin(Q) similar logic is used to show that the projection on the x plane is -v2cos(Q) therefore I end up with X(t)=-v2cos(Q)t + d, Y(t)=-v2sin(Q)t -(gt^2)/2+h . From here I have just been fidgeting with concepts and equations and have made no progress in showing that the inequality is true help would be appreciated. Also note that h is the height

 

[jbriggs444]

If you turn off gravity, how much time elapses before the bullets collide?
If you turn gravity back on, does that number change?

 

[Rubberduck2005]

jbriggs444 I am really stuck on this one could you give me any more than that?

 

[jbriggs444]

jbriggs444 I am really stuck on this one could you give me any more than that?

You have not answered either question yet.

 

[Frabjous]

Some observations
1) There are two conditions to determine when the bullets hit. You only need to use one of them.
2) There are two conditions to determine if the bullets hit above ground. You only need to use one of them.
3) d, h and Q are related so the final answer can look different depending on the decisions you make for 1 and 2. The final answer depends on d and Q which gives guidance for your choices for 1 and 2.
4) @jbriggs444 ‘s questions are trying to step you through this

 

[jbriggs444]

If you turn off gravity, how much time elapses before the bullets collide?
If you turn gravity back on, does that number change?

So let's take that first question. If we turn off gravity, we have two guns shooting bullets on a collision course. Both bullets follow the line of sight path between the two shooters.

If we knew the length of the line of sight path and the rate at which the two bullets were approaching each other, we might be able to use those pieces of information.

This hint was free. The next one will cost you more effort than that required to declare "I am stuck".

 

[PeroK]

Why can't one robber shoot the other in the head and the other shoot the first robber in the foot?

 

[Frabjous]

Why can't one robber shoot the other in the head and the other shoot the first robber in the foot?

We‘re physicists. Assume that the robbers are point particles.

 

[PeroK]

We‘re physicists. Assume that the robbers are point particles.

Shouldn't that be gun-point particles?

 

[Delta2]

Ok let's suppose for some reason you can't understand the hints you are given.

You have made some equations for $x(t),X(t),y(t),Y(t)$ though which seem correct to me. What is the relation that must hold between $x(t)$ and $X(t)$ and between $y(t)$ and $Y(t)$ so that the bullets meet at some time instance $t_0$?

What additional inequalities must hold for $y(t)$ and $Y(t)$ if the collision is to happen above ground?

 

[PeroK]

Ok let's suppose for some reason you can't understand the hints you are given.

You have made some equations for $x(t),X(t),y(t),Y(t)$ though which seem correct to me. What is the relation that must hold between $x(t)$ and $X(t)$ and between $y(t)$ and $Y(t)$ so that the bullets meet at some time instance $t_0$?

What additional inequalities must hold for $y(t)$ and $Y(t)$ if the collision is to happen above ground?

This is all right, but this problem is perhaps better tackled by considering different frames of reference.

 

[Delta2]

This is all right, but this problem is perhaps better tackled by considering different frames of reference.

Yes ok but maybe the OP has the same weakness as me, that is he is used to think in the "absolute" ground frame or lab frame.

Also trying to make myself follow the OP's way rather than make him follow my own "clever shortcut" way that he might not understand :P.

 

[jbriggs444]

Yes ok but maybe the OP has the same weakness as me, that is he is used to think in the "absolute" ground frame or lab frame.

Also trying to make myself follow the OP's way rather than make him follow my own "clever shortcut" way that he might not understand :P.

I think we lost OP a while back. *sob*, *sniff*.

 

[ergospherical]

Both bullets undergo equal acceleration so in the frame of the ground bullet the roof bullet approaches with constant velocity $-\mathbf{v}_1 + \mathbf{v}_2$. The time until collision is therefore $t_c = \dfrac{d}{(v_1 + v_2)\cos{Q}}$ which must be less than the time $t_* = \dfrac{2v_1 \sin{Q}}{g}$ for the first bullet to hit the ground.

 

[jbriggs444]

Both bullets undergo equal acceleration [...]

Yup. That is where I was trying to lead OP.