Projectile motion problem

Question

Two particles are projected with the same speed from the same point. The angles of projection are $2\alpha$ and $\alpha$ and a time $T$ elapses between the instants of projection. If the particles collide in flight, find the speed of projection in terms of $T$ and $\alpha$.

If the collision occurs when one of the particles is at its greatest height, show that $\alpha$ is given by $$4\cos^4 \alpha - \cos^2 \alpha -1 =0. $$

If the particles collide then $x_1=x_2$ where $x_1$ is the $x$-coordinate of the particle with projection angle 2α. We are given that they have they same speed at projection. The position of particle 1 is given by: $$x_1=u(t+T)\cos2α,$$ and $$y_1=u(t+T)\sin2α−\frac12g(t+T)^2.$$ Where t is the time taken for the second particle to hit the first. I understand that I need to solve for u, however I'm not sure of the best way to proceed. I can also find equations for the displacement for the first particle and similarly for the second particle. This all seems messy though!!

Answer 1

The two particles $p_1,p_2$ have the paths

$$ \cases{ p_1 = (v_0\cos(2\alpha)(t-T),v_0\sin(2\alpha)(t-T)-\frac 12 g(t-T)^2)\\ p_2 = (v_0\cos(\alpha)t,v_0\sin(\alpha)t-\frac 12 gt^2) } $$

at the intersection we have $p_1=p_2$ and eliminating $t$ we get

$$ v_0 = \frac 12 g(1-2\cos\alpha)\cot\left(\frac{\alpha}{2}\right)T $$

If the intersection occurs at the maximum height for the second particle, then we have to solve

$$ \cases{ p_1=p_2\\ t=\frac{v_0}{g}\sin\alpha } $$

and after eliminating $T$ we get

$$ \sin ^2\alpha (4 \cos ^4\alpha-\cos ^2\alpha-1)=0 $$

Answer 2

When $t=0$, the first particle is projected at angle $2\alpha$ and has displacements at time $t$ given by $$x_1=Vt\cos2\alpha$$ $$y_1=Vt\sin2\alpha-\frac12gt^2$$

The second particle is projected at angle $\alpha$ at time $t=T$ and has displacements at time $t$ given by

$$x_2=V(t-T)\cos\alpha$$ $$y_2=V(t-T)\sin\alpha-\frac12g(t-T)^2$$

They collide at time $t$, so setting $x_1=x_2$ gives $$t=\frac{T\cos\alpha}{\cos\alpha-\cos2\alpha}$$

Also, $$t-T=\frac{T\cos2\alpha}{\cos\alpha-\cos2\alpha}$$

Now setting $y_1=y_2$ gives, after cancelling $\frac12gt^2$, $$Vt\sin2\alpha=V(t-T)\sin\alpha+gtT-\frac12gT^2$$

Substituting for $t$ and $t-T$, $$\frac{VT\cos\alpha\sin2\alpha}{\cos\alpha-\cos2\alpha}=\frac{VT\cos2\alpha\sin\alpha}{\cos\alpha-\cos2\alpha}+\frac{gT^2\cos\alpha}{\cos\alpha-\cos2\alpha}-\frac12gT^2$$

This is easily simplified and rearranged to give

$$V=\frac{gT(\cos\alpha-\cos2\alpha)}{2\sin\alpha}$$

They collide at the maximum height of the second particle, so at time $t$, we have $$V\sin\alpha=g(t-T)$$

So now we can equate two expressions for $V$ and get, after cancelling $gT$,

$$\frac{\cos2\alpha}{(\cos\alpha-\cos2\alpha)\sin\alpha}=\frac{\cos\alpha+\cos2\alpha}{2\sin\alpha}$$

We can cancel $\sin\alpha$ since it is not zero, and the printed result follows in a couple of steps.