#### Answer

a). $The\, diagram \,is\, as\, shown.$
b). $680.4N$
c). $2.35s$

#### Work Step by Step

a). The diagram is as shown.
b). $F_{p}=m(g-a)$
Now, $h=0+\frac{1}{2}at^{2}$
$a=\frac{2h}{t^{2}}=\frac{2\times7}{2.5^{2}}=2.24 m/s^{2}$
Thus, $F_{p}=90(9.8-2.24)=680.4N$
c). $a=\frac{mg-F_{p}}{m}=\frac{75\times9.8-0.8\times680.4}{75}=2.54m/s^{2}$
$h=\frac{1}{2}at^{2}$
$t=\sqrt \frac{2h}{a}=\sqrt \frac{2\times7}{2.54}=2.35s$