A ball is dropped from the top of a tower. In the last second of motion it covers `(5)/(9)` th of tower height before hitting the ground. Acceleration due to gravity = 10 (in `ms^(–2))`. Find height of tower (in m):

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Motion of the Ball
The ball is dropped from the top of a tower and falls freely under the influence of gravity. The total height of the tower is denoted as \( h \).

### Step 2: Analyze the Last Second of Motion
In the last second of its motion, the ball covers \( \frac{5}{9} \) of the tower height. This means that during the last second, the distance traveled by the ball is:
\[ s_1 = \frac{5h}{9} \]
The remaining distance that the ball travels before the last second is:
\[ s_2 = h - s_1 = h - \frac{5h}{9} = \frac{4h}{9} \]

### Step 3: Define the Time of Fall
Let \( t \) be the total time taken for the ball to hit the ground. Therefore, the time taken to travel the distance \( s_2 \) is \( t - 1 \) seconds.

### Step 4: Apply the Equations of Motion
Using the equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
Since the ball is dropped (initial velocity \( u = 0 \)), the equation simplifies to:
\[ s = \frac{1}{2}gt^2 \]

1. For the total height \( h \):
\[ h = \frac{1}{2}gt^2 \]
2. For the distance \( s_2 \):
\[ \frac{4h}{9} = \frac{1}{2}g(t-1)^2 \]

### Step 5: Substitute Values into the Equations
Using \( g = 10 \, \text{m/s}^2 \):
1. From the first equation:
\[ h = \frac{1}{2} \cdot 10 \cdot t^2 = 5t^2 \]
2. From the second equation:
\[ \frac{4h}{9} = \frac{1}{2} \cdot 10 \cdot (t-1)^2 \]
Substituting \( h \) from the first equation into the second:
\[ \frac{4(5t^2)}{9} = 5(t-1)^2 \]
This simplifies to:
\[ \frac{20t^2}{9} = 5(t^2 - 2t + 1) \]
Dividing both sides by 5:
\[ \frac{4t^2}{9} = t^2 - 2t + 1 \]

### Step 6: Rearranging the Equation
Multiply through by 9 to eliminate the fraction:
\[ 4t^2 = 9t^2 - 18t + 9 \]
Rearranging gives:
\[ 5t^2 - 18t + 9 = 0 \]

### Step 7: Solve the Quadratic Equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 5, b = -18, c = 9 \):
\[ t = \frac{18 \pm \sqrt{(-18)^2 - 4 \cdot 5 \cdot 9}}{2 \cdot 5} \]
Calculating the discriminant:
\[ t = \frac{18 \pm \sqrt{324 - 180}}{10} = \frac{18 \pm \sqrt{144}}{10} = \frac{18 \pm 12}{10} \]
This gives two possible values for \( t \):
\[ t = \frac{30}{10} = 3 \quad \text{and} \quad t = \frac{6}{10} = 0.6 \, \text{(not valid since time cannot be less than 1)} \]
Thus, \( t = 3 \) seconds.

### Step 8: Calculate the Height of the Tower
Substituting \( t = 3 \) back into the equation for height:
\[ h = 5t^2 = 5 \cdot 3^2 = 5 \cdot 9 = 45 \, \text{m} \]

### Final Answer
The height of the tower is \( 45 \, \text{m} \).

---