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 step by step, we will use the equations of motion under uniform acceleration. Here's the detailed solution:

### Step 1: Understanding the Problem
A 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 acceleration due to gravity (g) on this planet.

### Step 2: Set Up the Equations
We will use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) = distance covered
- \( u \) = initial velocity (which is 0 since the ball is dropped)
- \( a \) = acceleration (which is \( g \))
- \( t \) = time

### Step 3: Distance Covered in the Last 0.5 Seconds
In the last 0.5 seconds before hitting the ground, the distance covered is 79 m. Thus, we can write:
\[ 79 = 0 \cdot (0.5) + \frac{1}{2} g (0.5)^2 \]
This simplifies to:
\[ 79 = \frac{1}{2} g \cdot \frac{1}{4} \]
\[ 79 = \frac{g}{8} \]
From this, we can express \( g \):
\[ g = 79 \cdot 8 = 632 \, \text{m/s}^2 \]

### Step 4: Total Distance Covered
Next, we need to find the total time \( T \) it takes to fall the entire height of 200 m. The ball falls for \( T - 0.5 \) seconds before the last 0.5 seconds.

Using the total distance equation:
\[ 200 = 0 \cdot T + \frac{1}{2} g T^2 \]
This simplifies to:
\[ 200 = \frac{1}{2} g T^2 \]
Substituting \( g = 632 \):
\[ 200 = \frac{1}{2} \cdot 632 \cdot T^2 \]
\[ 200 = 316 T^2 \]
\[ T^2 = \frac{200}{316} \]
\[ T^2 = 0.6329 \]
\[ T = \sqrt{0.6329} \approx 0.796 \, \text{s} \]

### Step 5: Calculate the Time Before Last 0.5 Seconds
The time before the last 0.5 seconds is:
\[ T - 0.5 \approx 0.796 - 0.5 = 0.296 \, \text{s} \]

### Step 6: Verify the Calculations
To ensure that the calculations are consistent, we can check the total distance covered in \( T \) seconds:
\[ s = \frac{1}{2} g T^2 \]
Substituting \( g \) and \( T \):
\[ s = \frac{1}{2} \cdot 632 \cdot (0.796)^2 \]
Calculating this should yield approximately 200 m.

### Final Answer
Thus, the acceleration due to gravity \( g \) on the planet is approximately:
\[ g \approx 632 \, \text{m/s}^2 \]

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