I'm stuck on the following simple projectile motion problem up a plane. Below is the question:
A plane is inclined at an angle $\alpha$ to the horizontal. A particle is projected up the plane with speed $u$ at an angle $\beta$ to the plane. The plane of projection is vertical and contains the line of greatest slope. When the particle is at its maximum perpendicular height above the plane, it is $\frac{3}{5}$ of the range up the plane. Show that $\tan\alpha \tan\beta = \frac{2}{7}$.
Here is what I did.
I set up a coordinate system parallel and perpendicular to the plane - $\vec x$ (up the plane) and $\vec y$ (perpendicular to the plane). Thus, I can express the initial velocity $u$ in terms if this coordinate system.
$$\vec u = u \cos\beta\vec x + u\sin\beta\vec y$$
Therefore
$$\vec u_x = u \cos\beta $$ $$\vec u_y = u \sin\beta $$
The acceleration due to gravity, in terms of $\vec x$ and $\vec y$ is given by:
$$ \vec a = -g\sin\alpha \vec x - g\cos\alpha \vec y$$
Therefore
$$\vec a_x = -g\sin\alpha $$ $$\vec a_y = -g\cos\alpha $$
For maximum perpendicular height ($h$) I used $$(\vec v_y)^2 = (\vec u_y)^2 + 2\vec a_y h$$ Since $\vec v_y = 0 $ for maximum height, I can find $h$.
$$ 0 = u^2\sin^2\beta + 2(-g\cos\alpha)h$$ $$ h = \frac{u^2\sin^2\beta}{2g\cos\alpha}$$
To find the range, I first found the time of flight. For the time of flight $\vec s_y = 0$. Using the formula:
$$ \vec s_y = \vec u_yt + \frac{1}{2} \vec a_y t^2$$ $$ 0 = u\sin \beta t + \frac{1}{2}(-g\cos \alpha) t^2 $$ $$ t(u\sin \beta - \frac{1}{2}g\cos \alpha t) = 0 $$
$$t = 0 , t = \frac{2u\sin \beta}{g\cos \alpha}$$
Time of flight
$$ t = \frac{2u\sin \beta}{g\cos \alpha} $$
To find the range I used the formula
$$ \vec s_x = \vec u_xt + \frac{1}{2} \vec a_x t^2$$ $$ \vec s_x = u \cos\beta \biggl(\frac{2u\sin \beta}{g\cos \alpha}\biggl) + \frac{1}{2} (-g\sin \alpha)\biggl(\frac{2u\sin \beta}{g\cos \alpha}\biggl)^2$$ $$ \vec s_x = \frac{2u^2\sin \beta \cos \beta}{g\cos \alpha} - \frac{2u^2 \sin \alpha \sin^2\beta}{g\cos^2 \alpha} $$
And this is where I get stuck. I can seem to derive $\tan\alpha \tan\beta = \frac{2}{7}$.
The hint is that the time taken to reach the maximum height is t/2, where t is the time of flight.
If you still cannot solve it, I have briefly written down the solution below.
$$s_{max\_ht} = \frac{u_xt}{2} + \frac{1}{2}.\frac{a_xt^2}{4} = \frac{3}{5}(u_xt + \frac{1}{2}a_xt^2)$$
$$\implies \frac{u_xt}{10} = \frac{a_xt^2}{2}.\frac{-7}{20}$$
Since you have taken $a_x$ as -g$\sin (\alpha)$, the negative sign gets cancelled and,
$$\frac{2u_x}{|a_x|t} = \frac{7}{2}$$
substituting the values of $u_x$, t, $|a_x|$ and taking the reciprocal, we get
$$tan(\alpha)tan(\beta) = \frac{2}{7}$$