Answer
a) $F_p=727N$
b) No
c) $F_p=931N$
Work Step by Step
(a) We know that
$F=W-F_p$
We plug in the known values to obtain:
$(102Kg)(2.67m/s^2)=(102Kg)(9.8m/s^2)-F_p$
This simplifies to:
$F_p=727N$
(b) We know that the acceleration of the fireman is directly proportional to the square of the landing speed. Thus, when the landing speed is halved, the acceleration of the fireman is reduced to four times. We conclude that when the landing speed is halved, the force exerted on the fireman cannot be doubled.
(c) We know that
$a=\frac{v^2-v_{\circ}^2}{2x}$
$\implies a=\frac{(2.1m/s)^2-(0m/s)^2}{2(3.3m)}0.668m/s^2$
Now $F_p=(102Kg)(9.8m/s^2-0.668m/s^2)$
$F_p=931N$
