An object falls through a distance h in certain time on the earth. The same object falls through a distance 4h in the same time on a planet. If 'g' is acceleration dur to gravity on the earth then acceleration due to gravity on that planet will be

Options

A. g/4
B. 4g
C. g/2
D. 2g

Explanation

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

### Step 1: Analyze the first case (on Earth)
When the object falls through a distance \( h \) under the influence of gravity \( g \), we can use the second equation of motion:

\[ h = \frac{1}{2} g t^2 \]

Where:
- \( h \) is the distance fallen (on Earth),
- \( g \) is the acceleration due to gravity on Earth,
- \( t \) is the time taken to fall that distance.

### Step 2: Rearranging the equation for time
From the equation above, we can solve for \( t^2 \):

\[ t^2 = \frac{2h}{g} \]

### Step 3: Analyze the second case (on the planet)
Now, consider the object falling through a distance \( 4h \) on the other planet with acceleration due to gravity \( g' \). Using the same equation of motion, we have:

\[ 4h = \frac{1}{2} g' t^2 \]

### Step 4: Substitute \( t^2 \) from the first case
Since the time \( t \) is the same for both cases, we can substitute \( t^2 \) from Step 2 into the equation for the second case:

\[ 4h = \frac{1}{2} g' \left(\frac{2h}{g}\right) \]

### Step 5: Simplifying the equation
Now, simplify the equation:

\[ 4h = \frac{1}{2} g' \cdot \frac{2h}{g} \]
\[ 4h = \frac{g' h}{g} \]

### Step 6: Isolate \( g' \)
To find \( g' \), multiply both sides by \( g \) and divide by \( h \):

\[ g' = 4g \]

### Final Answer
Thus, the acceleration due to gravity on the planet is:

\[ g' = 4g \]

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