Projectile Motion - Two fired at different times

[cnstntcnfsn]

Homework Statement

Two projectiles are fired, one at t = 0 with velocity v₀₁, making an angle θ₁ with x-axis, and the other at t = T > 0 with velocity v₀₂ making an angle θ₂ with x-axis. Show that if they are to collide in midair the travel interval T between the two firings must be:

$$\frac{2v_{01}v_{02}sin(θ_1-θ_2)}{g(v_{01}cos(θ_1) + v_{02}cos(θ_2))}$$

Homework Equations

$$sin(θ_1-θ_2) = sin(θ_1)cos(θ_2)-cos(θ_1)sin(θ_2)$$
$$\frac{d^2x}{dt^2} = 0, \frac{dx}{dt} = v_0x, x = v_{0x}t$$
$$\frac{d^2z}{dt^2} = -g, \frac{dz}{dt} = v_{0z} - gt, z = v_{0x}t - \frac{gt^2}{2}$$
$$\vec r = x \vec i + z \vec j = v_{0x}t \vec i + (v_{0x}t - \frac{gt^2}{2}) \vec j$$

The Attempt at a Solution

I used the above equations and the proper indices to create the following equations:

$$v_{01}cos(θ_1)t = v_{02}cos(θ_2)T$$
$$v_{01}sin(θ_1)t - \frac{gt^2}{2} = v_{02}sin(θ_2)T - \frac{gT^2}{2}$$

I then arranged these so that each equation equaled 0, and then equated them to each other. I got the following equation:

$$v_{01}cos(θ_1)t - v_{02}cos(θ_2)T = v_{01}sin(θ_1)t - \frac{gt^2}{2} - v_{02}sin(θ_2)T + \frac{gT^2}{2}$$

Simplifying it gave me:

$$v_{01}cos(θ_1)t - v_{02}cos(θ_2)T - v_{01}sin(θ_1)t + v_{02}sin(θ_2)T = \frac{g}{2} (T^2 - t^2)$$

$$\frac{2v_{01}v_{02}(t\frac{(cos(θ_1)-sin(θ_2))}{v_{02}})+T(\frac{(sin(θ_2)-cos(θ_2))}{v_{01}})}{g} = T^2 - t^2$$

I don't know where to go from here to get to the solution, or if my process is correct. I tried taking the norm of r and then equating |r₁| = |r₂|, but that just created a mess with a lot of terms that wouldn't cancel.

Thanks!

 

[cnh1995]

01sin(θ1)t−gt22=v02sin(θ2)T−gT2

You are asked to verify the expression for t-T. Rearrange this equation to separate t-T on one side. Then substitute t=x/v_o1cosθ1 and T=x/v_o2cosθ2 on the other side
and you'll get the desired expression.

 

[haruspex]

You seem to be constantly confused in this question about how to express the time information in the equations.

$v_{01}\cos(θ_1)t=v_{02}\cos(θ_2)T$

At time t > T, how long has the second particle been travelling?

 

[cnstntcnfsn]

You seem to be constantly confused in this question about how to express the time information in the equations.

At time t > T, how long has the second particle been travelling?

t-T

Would I then find t and T from x and plug that into z?

 

[cnh1995]

First particle travels for time t, second one travels for time T. You are asked to find the expression for t-T i.e. the interval between the two firings.

 

[haruspex]

First particle travels for time t, second one travels for time T. You are asked to find the expression for t-T i.e. the interval between the two firings.

No, you have it backwards. T is the interval between firings.

 

[haruspex]

t-T

Would I then find t and T from x and plug that into z?

Your first attempt was fine except that on the right hand side of your first two equations you had T where you should have had t-T. Try that again.

 

[cnstntcnfsn]

Your first attempt was fine except that on the right hand side of your first two equations you had T where you should have had t-T. Try that again.

Should my left side then be $t^2-T^2$?

 

[cnh1995]

No, you have it backwards. T is the interval between firings.

Assuming t and T to be the times of travel, I got the desired expression. Maybe I didn't read the qusation properly. But even if I assume so, it won't affect the final expression since there is no time term in it and I'm finding the difference t-T.

 

[cnh1995]

Should my left side then be $t^2-T^2$?

If you are assuming T to be the time difference between firings, you can replace t with t1 and T with t2 in your attempt. t1-t2 will be your T. So, your left hand side should be t1²-t2².

 

[haruspex]

Should my left side then be $t^2-T^2$?

Not sure what you are referring to. In your very first two equations of your attempt, you have t (only) on the left and T (only) on the right.
The left hand sides are fine; t is the time the first particle has been traveling at time t. Just replace all the Ts on the right with t-T, since that is the time the second particle has been traveling at time t.

 

[haruspex]

Assuming t and T to be the times of travel, I got the desired expression. Maybe I didn't read the qusation properly. But even if I assume so, it won't affect the final expression since there is no time term in it and I'm finding the difference t-T.

Sure, but it was going to be hellish confusing with different advice based on different definitions of the same symbols. Assuming cnstntcnfsn read the question correctly, you were outvoted.

 

[cnstntcnfsn]

Not sure what you are referring to. In your very first two equations of your attempt, you have t (only) on the left and T (only) on the right.
The left hand sides are fine; t is the time the first particle has been traveling at time t. Just replace all the Ts on the right with t-T, since that is the time the second particle has been traveling at time t.

I replaced them, but I am a little confused (I don't know why I can't get this as it seems simple). From the first equation I got $$\frac{v_{01}cos(\theta_1)t}{v_{02}cos(\theta_2)} = t - T$$
I then put this into the 2nd equation and got $$v_{01}sin(\theta_1)t = \frac{v_{02}sin(\theta_2)v_{01}cos(\theta_1)t}{v_{02}cos(\theta_2)}+g\frac{T^2}{2}$$
I can then get $$t(v_{01}sin(\theta_1)v_{02}cos(\theta_2)-v_{02}sin(\theta_2)v_{01}cos(\theta_1) )=t v_{01}v_{02}sin(\theta_1-\theta_2)= g\frac{T^2}{2}$$

Does that look on the right track?

 

[haruspex]

I replaced them, but I am a little confused (I don't know why I can't get this as it seems simple). From the first equation I got $$\frac{v_{01}cos(\theta_1)t}{v_{02}cos(\theta_2)} = t - T$$
I then put this into the 2nd equation and got $$v_{01}sin(\theta_1)t = \frac{v_{02}sin(\theta_2)v_{01}cos(\theta_1)t}{v_{02}cos(\theta_2)}+g\frac{T^2}{2}$$
I can then get $$t(v_{01}sin(\theta_1)v_{02}cos(\theta_2)-v_{02}sin(\theta_2)v_{01}cos(\theta_1) )=t v_{01}v_{02}sin(\theta_1-\theta_2)= g\frac{T^2}{2}$$

Does that look on the right track?

You have two unknowns, t and T, but you are only interested in T. So you should use one equation (better not the quadratic one) to express t as a function of T, then use that to get rid of t in the other equation.

 

[cnstntcnfsn]

You have two unknowns, t and T, but you are only interested in T. So you should use one equation (better not the quadratic one) to express t as a function of T, then use that to get rid of t in the other equation.

Got it! Thank you so much for your help