An astronaut is on the surface of other planet whose air resistance is negligible. To measure the acceleration due t o gravity `(g)`, he throws a stone upwards. He observer that the stone reaches to a maximum height of `10m` and reaches the surface `4` second after it was thrown. find the accelertion due to gravity `(g)` on the surface of that planet in `m//s^(2)`.

Explanation

To find the acceleration due to gravity \( g \) on the surface of the planet, we can use the information given in the problem. The stone reaches a maximum height of \( 10 \, \text{m} \) and takes a total of \( 4 \, \text{s} \) to return to the surface after being thrown upwards.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The stone is thrown upwards and reaches a maximum height of \( 10 \, \text{m} \).
- The total time taken for the stone to go up and come down is \( 4 \, \text{s} \).
- Therefore, the time taken to reach the maximum height (upward motion) is \( 2 \, \text{s} \) (half of the total time).

2. **Using the Kinematic Equation**:
- At the maximum height, the final velocity \( v = 0 \).
- We can use the kinematic equation:
\[ v^2 = u^2 - 2gH \]
where:
- \( v = 0 \) (final velocity at maximum height),
- \( u \) is the initial velocity,
- \( g \) is the acceleration due to gravity,
- \( H = 10 \, \text{m} \) (maximum height).

- Rearranging gives:
\[ 0 = u^2 - 2g \cdot 10 \]
\[ u^2 = 20g \quad \text{(Equation 1)} \]

3. **Using the Time of Flight**:
- We can also use the equation of motion for the upward journey:
\[ v = u - gt \]
- Substituting \( v = 0 \) and \( t = 2 \, \text{s} \):
\[ 0 = u - g \cdot 2 \]
\[ u = 2g \quad \text{(Equation 2)} \]

4. **Substituting Equation 2 into Equation 1**:
- Now, substitute \( u = 2g \) into Equation 1:
\[ (2g)^2 = 20g \]
\[ 4g^2 = 20g \]
- Dividing both sides by \( g \) (assuming \( g \neq 0 \)):
\[ 4g = 20 \]
\[ g = 5 \, \text{m/s}^2 \]

### Final Answer:
The acceleration due to gravity \( g \) on the surface of that planet is \( 5 \, \text{m/s}^2 \).

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