During the last second of its flight, a ball thrown vertically upwards covers one-half of the distan

Q: Why do we add the upward and downward distances instead of subtracting them?

The question specifies the 'distance covered', not the 'displacement'. Distance is the total path length traveled, so we must add the magnitude of the upward journey to the magnitude of the downward journey.

Q: How do we know the last second is entirely during the downward motion?

We are trying to maximize the total time of flight. If the last second included the peak, the total time would be very small (less than 2 seconds). By assuming the last second is purely downward, we allow for a much longer fall, which leads to the maximum possible duration.

Q: Why did we reject $t_{down} = 1$ second?

When we solved the derivative equation, we had the step $t_{down} - 2 = \sqrt{\dots}$. Since a square root must yield a non-negative result, $t_{down} - 2$ must be greater than or equal to zero. Therefore, $t_{down}$ must be at least 2, making $t_{down} = 1$ an extraneous root introduced by squaring both sides.

Q: Does the initial height of projection matter?

The exact initial height isn't explicitly needed to find the time, but the math implies it. Since $t_{up} = 1$s and $t_{down} = 3$s, the ball was thrown from a height of 40 meters. The problem allows for this by stating the point of projection and landing may not be on the same level.

Q: Can we solve this without calculus?

Yes, you can use the properties of quadratic equations or AM-GM inequalities, but calculus provides a straightforward and foolproof method to find the maximum of the function $T(t_{down})$.

Question

During the last second of its flight, a ball thrown vertically upwards covers one-half of the distance covered during the whole flight. The point of projection and the point of landing may or may not be in the same horizontal level. What maximum possible duration of the flight can be obtained? Neglect air resistance and assume acceleration of free fall to be $10 	ext{ m/s}^2$.

Sigma Percentile · Accepted Answer

### The Last Second Mystery: Maximizing Time of Flight

Imagine a ball thrown vertically upwards. It travels to a certain maximum height and then falls back down. The problem presents a fascinating, almost counter-intuitive condition: during the very last second of its flight, the ball covers exactly one-half of the total distance it covered during its entire journey. Our mission is to find the maximum possible duration of this flight.

### Visualizing the Journey

To tackle this, we must first break the flight into two distinct phases. Let $t_{up}$ be the time taken for the ball to reach its maximum height from the point of projection. Let $t_{down}$ be the time taken for the ball to fall from that maximum height to the ground. The total time of flight is simply $T = t_{up} + t_{down}$.

Because distance is a scalar quantity, the total distance $D$ covered by the ball is the sum of the upward path length and the downward path length. Using the kinematic equation $s = \frac{1}{2}at^2$ from the highest point (where velocity is zero), we can express these distances as:

$$d_{up} = \frac{1}{2}g t_{up}^2$$

$$d_{down} = \frac{1}{2}g t_{down}^2$$

Therefore, the total distance is $D = \frac{1}{2}g (t_{up}^2 + t_{down}^2)$.

### The Mathematical Translation

Now, let's focus on that crucial last second. To maximize the total time, it makes physical sense that this last second occurs entirely during the downward fall. The distance covered in this final second, $d_{last}$, is the difference between the distance fallen in time $t_{down}$ and the distance fallen in time $(t_{down} - 1)$:

$$d_{last} = \frac{1}{2}g t_{down}^2 - \frac{1}{2}g (t_{down} - 1)^2 = \frac{1}{2}g (2t_{down} - 1)$$

The core condition of the problem states that $d_{last} = \frac{D}{2}$. Substituting our expressions into this condition yields:

$$\frac{1}{2}g (2t_{down} - 1) = \frac{1}{2} \left[ \frac{1}{2}g (t_{up}^2 + t_{down}^2) ight]$$

Notice how elegantly the $\frac{1}{2}g$ terms cancel out. By multiplying the entire equation by 2 and rearranging, we can isolate $t_{up}$:

$$4t_{down} - 2 = t_{up}^2 + t_{down}^2$$

$$t_{up} = \sqrt{4t_{down} - t_{down}^2 - 2}$$

### The Calculus of Optimization

We now have the total time $T$ expressed as a function of a single variable, $t_{down}$:

$$T(t_{down}) = \sqrt{4t_{down} - t_{down}^2 - 2} + t_{down}$$

To find the maximum possible duration, we must differentiate this function with respect to $t_{down}$ and set the derivative equal to zero:

$$\frac{dT}{dt_{down}} = \frac{4 - 2t_{down}}{2\sqrt{4t_{down} - t_{down}^2 - 2}} + 1 = 0$$

$$\frac{2 - t_{down}}{\sqrt{4t_{down} - t_{down}^2 - 2}} = -1$$

This implies that $t_{down} - 2 = \sqrt{4t_{down} - t_{down}^2 - 2}$. Squaring both sides to eliminate the radical gives us a quadratic equation:

$$(t_{down} - 2)^2 = 4t_{down} - t_{down}^2 - 2$$

$$t_{down}^2 - 4t_{down} + 4 = 4t_{down} - t_{down}^2 - 2$$

$$2t_{down}^2 - 8t_{down} + 6 = 0 \implies t_{down}^2 - 4t_{down} + 3 = 0$$

Factoring this quadratic yields $(t_{down} - 1)(t_{down} - 3) = 0$, giving us two potential solutions: $t_{down} = 1$ or $t_{down} = 3$. However, looking back at our derivative step, we established that $t_{down} - 2$ must be positive for the square root to be valid. Thus, we must discard $t_{down} = 1$ as an extraneous root. The only physically valid solution for maximizing the time is **$t_{down} = 3 	ext{ s}$**.

### Final Calculation

Substituting $t_{down} = 3 	ext{ s}$ back into our equation for $t_{up}$:

$$t_{up} = \sqrt{4(3) - 3^2 - 2} = \sqrt{12 - 9 - 2} = 1 	ext{ s}$$

The maximum possible duration of the flight is the sum of these two times:

**$T_{max} = 1 	ext{ s} + 3 	ext{ s} = 4 	ext{ s}$**

This means the ball was thrown upwards, reached its peak in 1 second, and then fell for 3 seconds, covering exactly half of its total journey's distance in that final, thrilling second.

Enter Numerical Value:

Visualized Solution

$T = t_{u p} + t_{d o w n}$

$D = d_{u p} + d_{d o w n}$

$D = \frac{1}{2} g (t_{u p}^{2} + t_{d o w n}^{2})$

$d_{l a s t} = \frac{1}{2} g (2 t_{d o w n} - 1)$

$d_{l a s t} = \frac{D}{2}$

$2 t_{d o w n} - 1 = \frac{1}{2} (t_{u p}^{2} + t_{d o w n}^{2})$

$t_{u p} = 4 t_{d o w n} - t_{d o w n}^{2} - 2$

$T (t_{d o w n}) = 4 t_{d o w n} - t_{d o w n}^{2} - 2 + t_{d o w n}$

$\frac{d T}{d t _{d o w n}} = 0$

$t_{d o w n}^{2} - 4 t_{d o w n} + 3 = 0$

$t_{d o w n} = 3 s$

$T_{ma x} = 4 s$

The Sigma Insight: Motion in a Straight Line

The Last Second Mystery

Visualizing the Journey

The Mathematical Translation

The Calculus of Optimization

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=tup​+tdown​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D=dup​+ddown​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D=21​g(tup2​+tdown2​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dlast​=21​g(2tdown​−1)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dlast​=2D​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2tdown​−1=21​(tup2​+tdown2​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tup​=4tdown​−tdown2​−2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T(tdown​)=4tdown​−tdown2​−2​+tdown​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dtdown​dT​=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tdown2​−4tdown​+3=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tdown​=3 s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tmax​=4 s