A ball is dropped from the top of a 200 m high tower on a planet. In the last `1/2` s before hitting the ground, it covers a distance of 79 m. Acceleration due to gravity (in ms–2) near the surface on that planet is ______.

Explanation

To solve the problem, we need to find the acceleration due to gravity (g) on a planet where a ball is dropped from a height of 200 m and covers a distance of 79 m in the last 0.5 seconds before hitting the ground.

### Step-by-Step Solution:

1. **Understanding the Problem:**
The ball is dropped from a height of 200 m. In the last 0.5 seconds before it hits the ground, it covers a distance of 79 m. We need to find the value of g.

2. **Setting Up the Equations:**
Let \( t_0 \) be the time taken to reach the point where the ball has fallen 121 m (since \( 200 - 79 = 121 \)). The ball travels 121 m in time \( t_0 \) and 79 m in the last 0.5 seconds.

3. **Using the Second Equation of Motion:**
For the distance covered in the first part (121 m):
\[ s = ut + \frac{1}{2}gt^2 \]
Here, the initial velocity \( u = 0 \) (since the ball is dropped), so:
\[ 121 = 0 + \frac{1}{2}g(t_0)^2 \]
Simplifying gives:
\[ 121 = \frac{1}{2}g(t_0)^2 \implies g(t_0)^2 = 242 \implies t_0 = \sqrt{\frac{242}{g}} \]

4. **Distance Covered in the Last 0.5 seconds:**
For the last 0.5 seconds (79 m):
The total time to hit the ground is \( t_0 + 0.5 \). The distance covered in this time can also be calculated using the second equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
The total distance covered in time \( t_0 + 0.5 \):
\[ 200 = 0 + \frac{1}{2}g(t_0 + 0.5)^2 \]
Expanding this gives:
\[ 200 = \frac{1}{2}g(t_0^2 + t_0 + 0.25) \]
Rearranging gives:
\[ 400 = g(t_0^2 + t_0 + 0.25) \]

5. **Substituting for \( t_0^2 \):**
From the first equation, we have \( t_0^2 = \frac{242}{g} \). Substitute this into the second equation:
\[ 400 = g\left(\frac{242}{g} + t_0 + 0.25\right) \]
Rearranging gives:
\[ 400 = 242 + gt_0 + 0.25g \]
Thus:
\[ 400 - 242 = gt_0 + 0.25g \implies 158 = gt_0 + 0.25g \]

6. **Expressing \( t_0 \) in terms of g:**
From \( t_0 = \sqrt{\frac{242}{g}} \), substitute this into the equation:
\[ 158 = g\sqrt{\frac{242}{g}} + 0.25g \]
Simplifying gives:
\[ 158 = 15.554\sqrt{g} + 0.25g \]

7. **Solving for g:**
Rearranging and squaring both sides will yield a quadratic equation in terms of \( g \). After solving, we find:
\[ g \approx 79 \, \text{m/s}^2 \]

### Final Answer:
The acceleration due to gravity (g) near the surface of that planet is approximately **79 m/s²**.