A ball is dropped from the roof of a tower height `h`. The total distance covered by it in the last second of its motion is equal to the distance covered by it in first three seconds. The value of `h` in metre is `(g=10m//s^(2))`

Options

A. `125`
B. `200`
C. `100`
D. `80`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have a ball dropped from a height \( h \) and we need to find \( h \) given that the distance covered in the last second of its motion is equal to the distance covered in the first three seconds.

### Step 2: Use the equations of motion
The distance covered by an object under uniform acceleration can be calculated using the formula:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) is the distance,
- \( u \) is the initial velocity,
- \( a \) is the acceleration (which is \( g = 10 \, \text{m/s}^2 \) in this case),
- \( t \) is the time.

Since the ball is dropped, the initial velocity \( u = 0 \).

### Step 3: Calculate the total height \( h \)
The total distance covered by the ball when it hits the ground can be expressed as:
\[ h = \frac{1}{2} g t^2 \]
Substituting \( g = 10 \, \text{m/s}^2 \):
\[ h = \frac{1}{2} \times 10 \times t^2 = 5t^2 \quad \text{(1)} \]

### Step 4: Calculate the distance covered in the first three seconds
Using the same formula for the first three seconds:
\[ h' = \frac{1}{2} g (3^2) = \frac{1}{2} \times 10 \times 9 = 45 \, \text{m} \quad \text{(2)} \]

### Step 5: Calculate the distance covered in the last second
The distance covered in the last second can be calculated as:
\[ h' = \text{Distance covered in } (t-1) \text{ seconds} - \text{Distance covered in } (t-2) \text{ seconds} \]
Using the formula:
- Distance covered in \( t \) seconds:
\[ s_t = \frac{1}{2} g t^2 = 5t^2 \]
- Distance covered in \( t-1 \) seconds:
\[ s_{t-1} = \frac{1}{2} g (t-1)^2 = 5(t-1)^2 \]

Thus, the distance covered in the last second is:
\[ h' = s_t - s_{t-1} = 5t^2 - 5(t-1)^2 \]
Expanding \( (t-1)^2 \):
\[ h' = 5t^2 - 5(t^2 - 2t + 1) = 5t^2 - 5t^2 + 10t - 5 = 10t - 5 \quad \text{(3)} \]

### Step 6: Set the distances equal
According to the problem, the distance covered in the last second is equal to the distance covered in the first three seconds:
\[ 10t - 5 = 45 \]

### Step 7: Solve for \( t \)
Rearranging gives:
\[ 10t = 50 \implies t = 5 \, \text{s} \]

### Step 8: Substitute \( t \) back to find \( h \)
Now substitute \( t = 5 \) back into equation (1):
\[ h = 5t^2 = 5 \times 5^2 = 5 \times 25 = 125 \, \text{m} \]

### Final Answer
The height of the tower \( h \) is \( 125 \, \text{m} \).

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