Time interval btwn 2 particles colliding (each with diff v and elevation)

[cecejay]

Homework Statement

Two particles are projected from the same point with velocities v₁ and v₂ at elevation α1 and α2, respectively, (α1>α2). Show that if they collide in mid-air, interval between firings must be:

2v1v2sin(α1 - α2)/g(v1cosα1 + v2cosα2)

Homework Equations

y = xtanα - (1/2)(gx^2)(sec^2(α))/v^2

t = (vsinα)/g

The Attempt at a Solution

since both particles are colliding, they will have the same x,y value, therefore:

xtanα1 - (1/2)(gx^2)(sec^2(α1))/v1^2 = xtanα2 - (1/2)(gx^2)(sec^2(α2))/v2^2

xtanα1 - xtanα2 = (1/2)(gx^2)(sec^2(α2))/v2^2 - (1/2)(gx^2)(sec^2(α1))/v1^2

x(tanα1 - tanα2) = (1/2)(gx^2)[sec^2(α2)/v2^2 -sec^2(α1)/v1^2]

(tanα1 - tanα2) = (1/2)(gx)[sec^2(α2)/v2^2 -sec^2(α1)/v1^2]and with the other equation:

t1 = (v1sinα1)/g and t2 = (v2sinα2)/g

t1 - t1 = (v1sinα1)/g - (v2sinα2)/g

t1 - t1 = (1/g)[v1sinα1 - v2sinα2]

Now I'm stuck because I'm not sure how to plug one equation into the other... and by the same step eliminate the x from the equation.

Any clues would be appreciated! Thanks

 

[mark726]

!To eliminate the x from the equation, you can use the fact that the x values for both particles are equal at the time of collision. This means that you can set the equations for x from both particles equal to each other:

xtanα1 = xtanα2

Then, you can solve for x:

x = (v1sinα1)t1 = (v2sinα2)t2

Substituting this value for x into the equation you have for t1 - t2, you get:

(tanα1 - tanα2) = (1/g)[(v1sinα1)t1 - (v2sinα2)t2]

Now you can substitute this equation into the equation you have for (tanα1 - tanα2) in terms of t1 and t2:

(1/2)(gx)[sec^2(α2)/v2^2 -sec^2(α1)/v1^2] = (1/g)[(v1sinα1)t1 - (v2sinα2)t2]

Solving for t1 - t2, you get:

t1 - t2 = (2v1v2sin(α1 - α2))/g[(v1cosα1 + v2cosα2)]

And this is the same result as the one given in the forum post. So, the interval between firings must be:

2v1v2sin(α1 - α2)/g(v1cosα1 + v2cosα2)