A particle is dropped from height `h = 100 m`, from surface of a planet. If in last `1/2` sec of its journey it covers 19 m. Then value of acceleration due to gravity that planet is :

Options

A. `8 m/s^2`
B. `1/8 m/s^2`
C. `1/4 m/s^2`
D. `2 m/s^2`

Explanation

To solve the problem step by step, we will use the equations of motion under uniform acceleration.

### Step 1: Understand the problem
A particle is dropped from a height of \( h = 100 \, \text{m} \). We need to find the acceleration due to gravity \( g \) on the planet, given that the particle covers \( 19 \, \text{m} \) in the last \( \frac{1}{2} \) second of its fall.

### Step 2: Define the variables
Let:
- \( g \) = acceleration due to gravity on the planet (what we need to find)
- \( t \) = total time taken to fall from height \( h \)
- The distance covered in the last \( \frac{1}{2} \) second is \( s_{last} = 19 \, \text{m} \).

### Step 3: Use the equations of motion
1. The total distance fallen in time \( t \) is given by:
\[ h = \frac{1}{2} g t^2 \]
Therefore, we can write:
\[ 100 = \frac{1}{2} g t^2 \quad \text{(1)} \]

2. The distance fallen in the last \( \frac{1}{2} \) second can be calculated using the formula:
\[ s_{last} = s(t) - s(t - \frac{1}{2}) \]
where \( s(t) = \frac{1}{2} g t^2 \) and \( s(t - \frac{1}{2}) = \frac{1}{2} g (t - \frac{1}{2})^2 \).

Thus, we have:
\[ s_{last} = \frac{1}{2} g t^2 - \frac{1}{2} g (t - \frac{1}{2})^2 \]
Simplifying gives:
\[ s_{last} = \frac{1}{2} g \left[ t^2 - (t^2 - t + \frac{1}{4}) \right] = \frac{1}{2} g \left[ t - \frac{1}{4} \right] \]
Setting \( s_{last} = 19 \):
\[ 19 = \frac{1}{2} g \left[ t - \frac{1}{4} \right] \quad \text{(2)} \]

### Step 4: Solve the equations
From equation (1):
\[ g = \frac{200}{t^2} \quad \text{(3)} \]
Substituting equation (3) into equation (2):
\[ 19 = \frac{1}{2} \left(\frac{200}{t^2}\right) \left[ t - \frac{1}{4} \right] \]
This simplifies to:
\[ 19 = 100 \left(\frac{t - \frac{1}{4}}{t^2}\right) \]
Rearranging gives:
\[ 19 t^2 = 100 \left(t - \frac{1}{4}\right) \]
Expanding and rearranging:
\[ 19 t^2 = 100t - 25 \]
\[ 19 t^2 - 100t + 25 = 0 \]

### Step 5: Use the quadratic formula
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 19, b = -100, c = 25 \):
\[ t = \frac{100 \pm \sqrt{(-100)^2 - 4 \cdot 19 \cdot 25}}{2 \cdot 19} \]
Calculating the discriminant:
\[ 10000 - 1900 = 8100 \]
Thus,
\[ t = \frac{100 \pm 90}{38} \]
Calculating the two possible values:
1. \( t = \frac{190}{38} = 5 \)
2. \( t = \frac{10}{38} = \frac{5}{19} \) (not physically meaningful as time cannot be negative)

### Step 6: Calculate \( g \)
Substituting \( t = 5 \) into equation (3):
\[ g = \frac{200}{5^2} = \frac{200}{25} = 8 \, \text{m/s}^2 \]

### Final Answer
The acceleration due to gravity on the planet is \( g = 8 \, \text{m/s}^2 \).

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