Step 1: Set Up the Physical Model
Let the total flight be divided into two segments relative to the maximum height:
- Ascent Time ($t_1$): Time taken to reach the peak from the launch point. Distance traveled: $h_1 = \frac{1}{2}gt_1^2$.
- Descent Time ($t_2$): Time taken to fall from the peak to the ground. Distance traveled: $h_2 = \frac{1}{2}gt_2^2$.
Total Distance ($D$): sum of upward and downward distances.
$$D = h_1 + h_2 = \frac{g}{2}(t_1^2 + t_2^2)$$Total Duration ($T$):
$$T = t_1 + t_2$$Step 2: Apply the “Last Second” Constraint
The problem states: “Distance in the last second is half the total distance.”
The distance covered in the last second of a fall (from $t_2-1$ to $t_2$) is:
$$d_{last} = h(t_2) – h(t_2-1)$$ $$d_{last} = \frac{g}{2}t_2^2 – \frac{g}{2}(t_2-1)^2 = \frac{g}{2}(2t_2 – 1)$$Equating this to half the total distance ($D/2$):
$$\frac{g}{2}(2t_2 – 1) = \frac{1}{2} \left[ \frac{g}{2}(t_1^2 + t_2^2) \right]$$Canceling constants ($\frac{g}{2}$):
$$2t_2 – 1 = \frac{1}{2}(t_1^2 + t_2^2)$$ $$4t_2 – 2 = t_1^2 + t_2^2$$Solving for $t_1$ in terms of $t_2$:
Step 3: Define the Optimization Function
We want to maximize the total time $T$. Substitute $t_1$ into the time equation:
$$T(t_2) = t_2 + \sqrt{4t_2 – 2 – t_2^2}$$Step 4: Differentiate (Calculus from Image)
To find the maximum, take the derivative of $T$ with respect to $t_2$ and set it to zero:
$$\frac{dT}{dt_2} = \frac{d}{dt_2} \left( t_2 + (4t_2 – 2 – t_2^2)^{1/2} \right) = 0$$Using the chain rule:
$$1 + \frac{1}{2}(4t_2 – 2 – t_2^2)^{-1/2} \cdot (4 – 2t_2) = 0$$Rearranging the terms:
$$\frac{4 – 2t_2}{2\sqrt{4t_2 – 2 – t_2^2}} = -1$$ $$\frac{2 – t_2}{\sqrt{4t_2 – 2 – t_2^2}} = -1$$Cross-multiply and square both sides to remove the root and negative sign:
$$2 – t_2 = -\sqrt{4t_2 – 2 – t_2^2}$$ $$(2 – t_2)^2 = 4t_2 – 2 – t_2^2$$ $$4 – 4t_2 + t_2^2 = 4t_2 – 2 – t_2^2$$Step 5: Solve the Quadratic Equation
Move all terms to one side:
$$2t_2^2 – 8t_2 + 6 = 0$$Divide by 2:
$$t_2^2 – 4t_2 + 3 = 0$$Factorize:
$$(t_2 – 3)(t_2 – 1) = 0$$This gives two possible values for the descent time:
$t_2 = 1$ or $t_2 = 3$
Step 6: Final Verification
Calculate the total time $T$ for both cases to find the maximum.
Case 1: If $t_2 = 1$
$$t_1 = \sqrt{4(1) – 2 – (1)^2} = \sqrt{1} = 1$$ $$T = 1 + 1 = 2 \text{ seconds}$$Case 2: If $t_2 = 3$
$$t_1 = \sqrt{4(3) – 2 – (3)^2} = \sqrt{12 – 2 – 9} = \sqrt{1} = 1$$ $$T = 1 + 3 = 4 \text{ seconds}$$