A ball is released from the top of a tower of height h metre. It takes T second to reach the ground.

Q: Why is the initial velocity zero?

Because the ball is 'released' or 'dropped', not thrown downwards. When an object is just let go, its starting velocity is $0$.

Q: Why did we use the second equation of motion?

Because it directly relates displacement ($s$), initial velocity ($u$), time ($t$), and constant acceleration ($a$). We have information about time and need to find displacement.

Q: I got $h/9$, why is it wrong?

You found the distance fallen from the top. The question specifically asks for the position from the ground, which is the total height minus the fallen distance ($h - h/9$).

Q: Does the mass of the ball matter?

No, acceleration due to gravity $g$ is independent of mass. All objects fall at the same rate in a vacuum.

Q: Can we use $v^2 = u^2 + 2as$?

We don't know the final velocities at those specific times, so using the third equation would require extra steps to find the velocities first. The second equation is much more direct.

Question

A ball is released from the top of a tower of height $h$ metre. It takes $T$ second to reach the ground. What is the position of the ball in $T /3$ second?

Options:

(A) $h / 9$ m from the ground

(B) $7h / 9$ m from the ground

(C) $8h / 9$ m from the ground

(D) $17h / 18$ m from the ground

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine standing at the top of a tall tower of height $h$. You hold a ball and simply let it go. Since you just release it, its initial velocity is zero ($u = 0$).

To find out how far the ball falls in a given time, we need a relationship between distance, time, and acceleration. The perfect tool for this is Newton's second equation of motion: 
$$s = ut + \frac{1}{2}at^2$$

### The Master Equation
We are told it takes exactly $T$ seconds to hit the ground. Let's plug this into our equation. The total distance is $h$, initial velocity is zero, and acceleration is $g$.

$$h = 0 \cdot T + \frac{1}{2}gT^2$$
$$h = \frac{1}{2}gT^2$$

This gives us our master equation for the total height.

### Finding the Partial Fall
Now, where is the ball at one-third of the total time? Let's call the distance it has fallen from the top $s$. We substitute $t = \frac{T}{3}$ into our equation.

$$s = 0 \cdot \left(\frac{T}{3}ight) + \frac{1}{2}g\left(\frac{T}{3}ight)^2$$

Let's carefully square the time term. $\frac{T}{3}$ squared becomes $\frac{T^2}{9}$.

$$s = \frac{1}{2}g \left(\frac{T^2}{9}ight)$$

Notice how we can pull the $\frac{1}{9}$ out to the front. Look closely at what remains inside the bracket.

$$s = \frac{1}{9} \left(\frac{1}{2}gT^2ight)$$

Do you recognize the term $\frac{1}{2}gT^2$? Yes, that's exactly our total height $h$! Substituting $h$ back in, we find that the ball has fallen a distance of $\frac{h}{9}$ from the top.

### Final Calculation
**There is a catch here!** The question asks for the position from the ground, not from the top. So, we must subtract the fallen distance from the total height.

$$	ext{Position from ground} = h - s$$
$$	ext{Position from ground} = h - \frac{h}{9} = \frac{8h}{9}$$

That's our final answer.

As a bonus, this is a beautiful application of **Galileo's law of odd numbers**. If you divide the total time into three equal intervals, the distances fallen are in the ratio $1 : 3 : 5$. The first interval covers $\frac{1}{9}$th of the total height. Think about how you can use this trick to solve similar problems in seconds!

Animated Solution for Physics - Kinematics: A ball is released from the top of a tower of height $h$ metre. It takes $T$ second to reach the ground. What is the position of the ball in $T /3$ second?

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Visualized Solution

$Visualizing the Fall$

$Equation of Motion$

$Total Time of Fall$

$Distance in t = T /3$

$Simplifying the Expression$

$Substituting h$

$Position from the Ground$

$Galileo’s Law of Odd Numbers$

The Sigma Insight: Motion in a Straight Line

Analyzing the Setup Imagine standing at the top of a tall tower of height $h$

The Master Equation We are told it takes exactly $T$ seconds to hit the ground

Finding the Partial Fall Now, where is the ball at one-third of the total time? Let's call the distance it has fallen from the top $s$

Final Calculation There is a catch here! The question asks for the position from the ground, not from the top

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Fall

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of Motion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Time of Fall

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Distance in t=T/3

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Expression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting h

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Position from the Ground

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Galileo’s Law of Odd Numbers