On a planet a ball is dropped from the top of a 100 m high tower. In the last `(1)/(2)s` before hitting the ground, it covers a distance of 19 m. Acceleration due to gravity (in `ms^(-2)` ) near the surface on that planet is _________ .

Explanation

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

### Step 1: Understand the problem
A ball is dropped from a height of 100 m. The ball covers 19 m in the last 0.5 seconds before hitting the ground. We need to find the acceleration due to gravity (g) on that planet.

### Step 2: Define variables
- Let the total time of flight be \( T \) seconds.
- The height of the tower \( H = 100 \) m.
- The distance covered in the last \( \frac{1}{2} \) seconds is \( S = 19 \) m.

### Step 3: Use the first equation of motion
The distance covered by the ball in \( T \) seconds can be given by:
\[ H = ut + \frac{1}{2} g T^2 \]
Since the ball is dropped, the initial velocity \( u = 0 \):
\[ 100 = 0 + \frac{1}{2} g T^2 \]
This simplifies to:
\[ 100 = \frac{1}{2} g T^2 \quad \text{(Equation 1)} \]

### Step 4: Calculate the distance covered in \( T - \frac{1}{2} \) seconds
The distance covered in \( T - \frac{1}{2} \) seconds is:
\[ S' = u(T - \frac{1}{2}) + \frac{1}{2} g (T - \frac{1}{2})^2 \]
Again, since \( u = 0 \):
\[ S' = \frac{1}{2} g (T - \frac{1}{2})^2 \]
We know that \( S' = 100 - 19 = 81 \) m, so:
\[ 81 = \frac{1}{2} g (T - \frac{1}{2})^2 \quad \text{(Equation 2)} \]

### Step 5: Solve for \( g \) using both equations
From Equation 1:
\[ g = \frac{200}{T^2} \]

Substituting \( g \) in Equation 2:
\[ 81 = \frac{1}{2} \left(\frac{200}{T^2}\right) \left(T - \frac{1}{2}\right)^2 \]
This simplifies to:
\[ 162 = \frac{200}{T^2} \left(T - \frac{1}{2}\right)^2 \]
Rearranging gives:
\[ 162 T^2 = 200 \left(T - \frac{1}{2}\right)^2 \]

### Step 6: Expand and simplify
Expanding the right side:
\[ 162 T^2 = 200 \left(T^2 - T + \frac{1}{4}\right) \]
\[ 162 T^2 = 200 T^2 - 200 T + 50 \]
Bringing all terms to one side:
\[ 0 = 38 T^2 - 200 T + 50 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 38 \), \( b = -200 \), \( c = 50 \):
\[ T = \frac{200 \pm \sqrt{(-200)^2 - 4 \cdot 38 \cdot 50}}{2 \cdot 38} \]
Calculating the discriminant:
\[ = \frac{200 \pm \sqrt{40000 - 7600}}{76} = \frac{200 \pm \sqrt{32400}}{76} = \frac{200 \pm 180}{76} \]
Calculating the two possible values for \( T \):
1. \( T = \frac{380}{76} = 5 \) seconds
2. \( T = \frac{20}{76} \) (not physically meaningful since it is less than 1 second)

### Step 8: Substitute \( T \) back to find \( g \)
Using \( T = 5 \) in Equation 1:
\[ 100 = \frac{1}{2} g (5^2) \]
\[ 100 = \frac{1}{2} g (25) \]
\[ g = \frac{200}{25} = 8 \, \text{m/s}^2 \]

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