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:

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

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

And so:

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

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

And:

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

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

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

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

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

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

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

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

We are told that we must have:

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

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

This simplifies to:

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

Divide through by $\cos(b)$:

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

And this reduces to:

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