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. What should be the initial downward speed of the rocketeer so that he catches the student at the top of 100 the floor for safe landing ?

Options

A. A. It can have many values
B. B. `180 ms^(-1)`
C. C. `175 ms^(-1)`
D. D. Cannot be determined

Explanation

To solve the problem step by step, we will analyze the situation involving the student and the rocketeer.

### Step 1: Determine the height from which the student jumps
The building has 145 floors, and each floor is 4 meters high. The height of the building can be calculated as:
\[ \text{Total height} = \text{Number of floors} \times \text{Height of each floor} = 145 \times 4 = 580 \text{ meters} \]

### Step 2: Calculate the height when the student reaches the top of the 100th floor
The student falls freely and reaches the top of the 100th floor:
\[ \text{Height of 100th floor} = 100 \times 4 = 400 \text{ meters} \]

### Step 3: Determine the time taken by the student to fall to the 100th floor
The stopwatch reads 6 seconds when the student reaches the top of the 100th floor. Thus, the time taken by the student to fall from the top of the building to the 100th floor is 6 seconds.

### Step 4: Calculate the distance fallen by the student
The distance fallen by the student is:
\[ \text{Distance fallen} = \text{Total height} - \text{Height of 100th floor} = 580 - 400 = 180 \text{ meters} \]

### Step 5: Calculate the average speed of the student
The average speed of the student can be calculated using the formula:
\[ \text{Average speed} = \frac{\text{Distance}}{\text{Time}} = \frac{180 \text{ m}}{6 \text{ s}} = 30 \text{ m/s} \]

### Step 6: Determine the time taken by the rocketeer to reach the same height
The rocketeer arrives 5 seconds after the student starts falling, which means he has only 1 second to reach the top of the 100th floor (since the student took 6 seconds to reach there).

### Step 7: Use the equations of motion to find the initial speed of the rocketeer
We can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \(s = 180 \text{ m}\) (distance fallen by the rocketeer),
- \(u\) is the initial speed of the rocketeer,
- \(t = 1 \text{ s}\),
- \(a = g = 10 \text{ m/s}^2\) (acceleration due to gravity).

Substituting the values into the equation:
\[ 180 = u \cdot 1 + \frac{1}{2} \cdot 10 \cdot (1^2) \]
\[ 180 = u + 5 \]
\[ u = 180 - 5 = 175 \text{ m/s} \]

### Conclusion
The initial downward speed of the rocketeer should be **175 m/s** in order to catch the student at the top of the 100th floor for a safe landing.