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.
In Q.1, what would be the approximate retardation to be given by jet pack along for safe landing?

Options

A. `5 g ms^(-2)`
B. `2 g ms^(-2)`
C. `4 g ms^(-2)`
D. Cannot be determined

Explanation

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

### Step 1: Calculate the height of the building
The height of the building can be calculated using the number of floors and the height of each floor.
- Number of floors = 145
- Height of each floor = 4 m

\[ \text{Total height} = \text{Number of floors} \times \text{Height of each floor} = 145 \times 4 = 580 \, \text{m} \]

**Hint:** Remember that the total height is the product of the number of floors and the height of each floor.

### Step 2: Determine the time taken by the student to fall
The student jumps off the building with negligible speed (initial velocity \( u = 0 \)). The time taken for the student to fall can be calculated using the equation of motion:

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

Where:
- \( s = 580 \, \text{m} \) (height of the building)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( u = 0 \)

Rearranging the equation gives:

\[ 580 = 0 + \frac{1}{2} \cdot 9.8 \cdot t^2 \]

\[ 580 = 4.9t^2 \]

\[ t^2 = \frac{580}{4.9} \approx 118.37 \]

\[ t \approx \sqrt{118.37} \approx 10.9 \, \text{s} \]

**Hint:** Use the equation of motion to relate distance, initial velocity, time, and acceleration.

### Step 3: Calculate the velocity of the student just before the rocketeer jumps
The time taken by the rocketeer to reach the student is 5 seconds after the student jumps. Therefore, the total time the student has been falling when the rocketeer jumps is:

\[ t_{total} = 5 + t_{rocketeer} \]

Using the time calculated earlier, we find that the student has fallen for approximately \( 10.9 \, \text{s} \).

Now, we can find the velocity of the student just before the rocketeer catches him:

\[ v = u + gt \]

Where \( u = 0 \):

\[ v = 0 + 9.8 \cdot 10.9 \approx 106.42 \, \text{m/s} \]

**Hint:** The velocity of a freely falling object can be calculated using the initial velocity, acceleration, and time.

### Step 4: Determine the conditions for the rocketeer to catch the student
The rocketeer jumps with an initial downward speed \( v_0 \) and must catch the student at a height where both can safely land. The rocketeer must have a final velocity of zero upon landing.

Using the equation of motion again:

\[ v^2 = u^2 + 2as \]

Setting \( v = 0 \) (final velocity), we have:

\[ 0 = v_0^2 + 2(-a)s \]

Where \( a \) is the retardation provided by the jet pack, and \( s \) is the distance fallen after catching the student.

**Hint:** Remember that the final velocity must be zero for a safe landing, and use the equation of motion to relate the velocities, acceleration, and distance.

### Step 5: Calculate the maximum allowable acceleration
The problem states that the magnitude of the acceleration (downward) should not exceed \( 5g \). Therefore, the total downward acceleration when the rocketeer catches the student must be:

\[ g - a \leq 5g \]

Rearranging gives:

\[ a \geq -4g \]

Thus, the retardation provided by the jet pack must be at least \( 4g \) (upward) to ensure a safe landing.

**Hint:** Use the conditions given in the problem to set up inequalities for the accelerations involved.

### Final Answer
The approximate retardation to be given by the jet pack for a safe landing is \( 4g \) (where \( g \approx 9.8 \, \text{m/s}^2 \)), which is approximately \( 39.2 \, \text{m/s}^2 \).