An inquistive student, determined to test the law of gravity for himself, walks to the top of a building og 145 floors, with every floor of height 4 m, having a stopwatch in his hand (the first floor is at a height of 4 m from the ground level). From there he jumps off with negligible speed and hence starts rolling freely. A rocketeer arrives at the scene 5 s later and dives off from the top of the building to save the student. The rocketeer leaves the roof with an initial downward speed `v_0`. In order to catch the student at a sufficiently great height above ground so that the rocketeer and the student slow down and arrive at the ground with zero velocity. The upward acceleration that accomplishes this is provided by rocketeer's jet pack, which he turns on just as he catches the student, before the rocketeer is in free fall. To prevent any discomfort to the student, the magnitude of the acceleration of the rocketeer and the student as they move downward together should not exceed 5 g.
Just as the student starts his free fall, he presses the button of the stopwatch. When he reaches at the top of 100th floor, he has observed the reading of stopwatch as 00:00:06:00 `(hh:mm:ss:100 th part of the second). Find the value of g. (correct upt ot two decimal places).

Options

A. `10.00 ms^(-2)`
B. `9.25 ms^(-2)`
C. `9.75 ms^(-2)`
D. `9.50 ms^(-2)`

Explanation

To solve the problem, we need to determine the value of \( g \) (acceleration due to gravity) based on the student's free fall from the building. Let's break this down step by step.

### Step 1: Calculate the height of the fall
The height of the building is given as 145 floors, with each floor having a height of 4 meters. Thus, the total height \( H \) of the building can be calculated as:

\[ H = \text{Number of floors} \times \text{Height of each floor} = 145 \times 4 = 580 \text{ meters} \]

### Step 2: Determine the distance fallen by the student
The student falls from the top of the building and we are interested in the distance he falls until he reaches the 100th floor. The height of the 100th floor from the ground is:

\[ \text{Height of 100th floor} = 100 \times 4 = 400 \text{ meters} \]

The distance \( s \) fallen by the student from the top of the building to the 100th floor is:

\[ s = H - \text{Height of 100th floor} = 580 - 400 = 180 \text{ meters} \]

### Step 3: Use the equation of motion
The student falls freely under gravity, so we can use the second equation of motion:

\[ s = ut + \frac{1}{2} a t^2 \]

Where:
- \( s \) = distance fallen = 180 m
- \( u \) = initial velocity = 0 m/s (since he jumps with negligible speed)
- \( t \) = time taken = 6 s
- \( a \) = acceleration due to gravity \( g \)

Substituting the known values into the equation:

\[ 180 = 0 \cdot 6 + \frac{1}{2} g (6^2) \]

This simplifies to:

\[ 180 = \frac{1}{2} g \cdot 36 \]

### Step 4: Solve for \( g \)
Now, rearranging the equation to solve for \( g \):

\[ 180 = 18g \]

Dividing both sides by 18:

\[ g = \frac{180}{18} = 10 \text{ m/s}^2 \]

### Conclusion
The value of \( g \) calculated based on the student's observations is:

\[ g = 10 \text{ m/s}^2 \]