MHB Projectile Fired: Find Time of Flight at Max Height

[markosheehan]

a plane is inclined at an angle a to the horizontal . a particle up the plane with speed u at an angle b to the plane . when the particle is at its max perpendicular height above the plane it is 3/5 of the range up the plane. show tana*tanb=2/7

so far I have found the time of flight which is 2usinb/gcosa

 

[MarkFL]

I would orient my coordinate axes such that the $x$-axis coincides with the plane, and the origin coincides with the initial position of the particle and so we have:

$$a_x=-g\sin(a)$$

$$a_y=-g\cos(a)$$

And so:

$$v_x=-g\sin(a)t+u\cos(b)$$

$$v_y=-g\cos(a)t+u\sin(b)$$

And:

$$x=-\frac{g\sin(a)}{2}t^2+u\cos(b)t$$

$$y=-\frac{g\cos(a)}{2}t^2+u\sin(b)t$$

When the particle has reached its range up the plane, we have $y=0$ and $t>0$ giving us:

$$t=\frac{2u\sin(b)}{g\cos(a)}$$

And so the range $R$ of the particle is:

$$R=\left(\frac{2u^2\sin(b)}{g\cos(a)}\right)\left(\cos(b)-\tan(a)\sin(b)\right)$$

When the particle has reached its maximum height, we find from $v_y=0$:

$$t=\frac{u\sin(b)}{g\cos(a)}$$

We are told that we must have:

$$x\left(\frac{u\sin(b)}{g\cos(a)}\right)=\frac{3}{5}R$$

$$-\frac{g\sin(a)}{2}\left(\frac{u\sin(b)}{g\cos(a)}\right)^2+u\cos(b)\left(\frac{u\sin(b)}{g\cos(a)}\right)=\frac{3}{5}\left(\frac{2u^2\sin(b)}{g\cos(a)}\right)\left(\cos(b)-\tan(a)\sin(b)\right)$$

This simplifies to:

$$2\cos(b)-\tan(a)\sin(b)=\frac{12}{5}\left(\cos(b)-\tan(a)\sin(b)\right)$$

Divide through by $\cos(b)$:

$$2-\tan(a)\tan(b)=\frac{12}{5}\left(1-\tan(a)\tan(b)\right)$$

And this reduces to:

$$\tan(a)\tan(b)=\frac{2}{7}$$