A body is dropped from a tower. It covers `64%` distance of its total height in last second. Find out the height of tower `[g=10 ms^(-2)]`

Answer

To solve the problem step by step, we can follow these calculations:

### Step 1: Understanding the Problem
A body is dropped from a tower and covers 64% of the total height in the last second. We need to find the height of the tower (h).

### Step 2: Define Variables
Let:
- Total height of the tower = \( h \)
- Distance covered in the last second = \( 0.64h \)
- Distance covered in the first \( t-1 \) seconds = \( 0.36h \)

### Step 3: Use Kinematic Equations
1. **Distance covered in the last second**:
The distance covered in the last second can be expressed using the formula:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( u = 0 \) (initial velocity), \( a = g = 10 \, \text{m/s}^2 \), and for the last second, the time is \( t - 1 \):
\[ 0.36h = 0 + \frac{1}{2} \cdot 10 \cdot (t - 1)^2 \]
Simplifying this gives:
\[ 0.36h = 5(t - 1)^2 \quad \text{(Equation 1)} \]

2. **Total distance covered**:
The total distance covered in \( t \) seconds is:
\[ h = 0 + \frac{1}{2} \cdot 10 \cdot t^2 \]
Simplifying this gives:
\[ h = 5t^2 \quad \text{(Equation 2)} \]

### Step 4: Relate Equations
From Equation 1, we have:
\[ 0.36h = 5(t - 1)^2 \]
Substituting \( h \) from Equation 2 into this equation:
\[ 0.36(5t^2) = 5(t - 1)^2 \]
This simplifies to:
\[ 1.8t^2 = 5(t - 1)^2 \]

### Step 5: Expand and Rearrange
Expanding the right side:
\[ 1.8t^2 = 5(t^2 - 2t + 1) \]
This gives:
\[ 1.8t^2 = 5t^2 - 10t + 5 \]
Rearranging terms:
\[ 0 = 5t^2 - 1.8t^2 - 10t + 5 \]
\[ 0 = 3.2t^2 - 10t + 5 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3.2 \), \( b = -10 \), and \( c = 5 \):
\[ t = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3.2 \cdot 5}}{2 \cdot 3.2} \]
Calculating the discriminant:
\[ = \frac{10 \pm \sqrt{100 - 64}}{6.4} \]
\[ = \frac{10 \pm \sqrt{36}}{6.4} \]
\[ = \frac{10 \pm 6}{6.4} \]
Calculating the two possible values for \( t \):
1. \( t = \frac{16}{6.4} = 2.5 \)
2. \( t = \frac{4}{6.4} = 0.625 \) (not valid since time cannot be less than 1 second)

### Step 7: Calculate Height
Using \( t = 2.5 \) seconds in Equation 2:
\[ h = 5t^2 = 5(2.5)^2 = 5 \cdot 6.25 = 31.25 \, \text{m} \]

### Final Answer
The height of the tower is \( h = 31.25 \, \text{meters} \).

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