y = P t 2 - Q t 3   m is the vertical displacement of a ball which is moving in vertical plane. Then…

Velocity is written as the rate of change of displacement.

$$\begin{gathered} v_y=\frac{\mathrm{d}y}{\mathrm{d}t} \\ =\frac{\mathrm{d}\left(P{t}^2-Q{t}^3\right)}{\mathrm{d}t} \\ =2Pt-3Q{t}^2 \end{gathered}$$

At maximum height, the velocity of the particle will be zero. Therefore,

$$\begin{gathered} t(2P-3Qt)=0 \\ \Rightarrow t=0 \& \frac{2P}{3Q} \end{gathered}$$

Considering $t=\frac{2P}{3Q}$ as $t=0$ means begining of motion.

Now maximum height will be,

$$\begin{gathered} y=P{t}^2-Q{t}^3 \\ =P{\left(\frac{2P}{3Q}\right)}^2-Q{\left(\frac{2P}{3Q}\right)}^3 \\ ={\left(\frac{2P}{3Q}\right)}^2\left(P-\frac{2P}{3}\right) \\ =\frac{4P^3}{27Q^2} \end{gathered}$$