Step 1: Understand the displacement function.
The vertical displacement is $y = Pt^2 - Qt^3$. The ball rises, slows, and reaches a top point; at that top the vertical velocity is momentarily zero. So our plan is to find the time of zero velocity and plug it back.
Step 2: Differentiate to get the velocity.
\[ v = \frac{dy}{dt} = 2Pt - 3Qt^2 \]
Step 3: Set the velocity to zero for maximum height.
\[ 2Pt - 3Qt^2 = 0 \] \[ t(2P - 3Qt) = 0 \] So either $t = 0$ (the start) or $2P - 3Qt = 0$. The useful root is \[ t = \frac{2P}{3Q} \]
Step 4: Substitute this time into $y$.
\[ y_{max} = P\left(\frac{2P}{3Q}\right)^2 - Q\left(\frac{2P}{3Q}\right)^3 \]
Step 5: Simplify each term.
First term: $P \cdot \dfrac{4P^2}{9Q^2} = \dfrac{4P^3}{9Q^2}$. Second term: $Q \cdot \dfrac{8P^3}{27Q^3} = \dfrac{8P^3}{27Q^2}$. So \[ y_{max} = \frac{4P^3}{9Q^2} - \frac{8P^3}{27Q^2} \]
Step 6: Combine over a common denominator.
Writing the first term as $\dfrac{12P^3}{27Q^2}$ gives \[ y_{max} = \frac{12P^3 - 8P^3}{27Q^2} = \frac{4P^3}{27Q^2} \] which is option (3). \[ \boxed{\dfrac{4P^3}{27Q^2}} \]