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

Answer

C

Explanation

c. Let time taken by rocketeer in
catching the boy at the top of 100th
floor be t, then time for which boy is in free fall motion is `(t+5)`s. So
`t+5 = 6 rArr t = 1s`
Now `180 = v_0t + ("gt"^2)/2`.
where `v_0` is the initial downward speed of racketeer.
which gives `v_0 = 175 ms ^(-1)` .