(a) Here, height(h)=1 km = 1000m, g = 10m/s²
Velocity attained by the rain drop in freely falling through a height h.
\(v=\sqrt { 2gh } =\sqrt { 2\times 10\times 1000 }\)
\( \\ =100\sqrt { 2 } m/s\)
\(\\ =100\sqrt { 2 } \times \frac { 60\times 60 }{ 1000 } Km/h\)
\(\\ =360\sqrt { 2 } km/h\approx 510km/h\) 
(b) Diameter of the drop(d) = 2r = 4mm
Radius of the drop(r) = 2mm = 2 x 10^-3m
Mass of a rain drop(m)=Vx\(\rho \)
\(=\frac { 4 }{ 3 } \pi { r }^{ 3 }\rho =\frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times (2\times { 10 }^{ -3 })^{ 3 }\times { 10 }^{ 3 }\)
\( \approx 3.4\times { 10 }^{ -5 }kg\)
\(\\ Momentum \ of \ the \ rain \ drop(\rho )=mv \)
\(\\ =3.4\times { 10 }^{ -5 }\times 100\sqrt { 2 }\)
\(\\ \approx 4.7\times { 10 }^{ -3 }{ kg }^{ -m/s }\)
(c) Time required to flatten the drop
= time taken by the drop to travel the distance equal to the diameter of the drop near the ground
\(t=\frac { d }{ v } =\frac { 4\times { 10 }^{ -3 } }{ 100\sqrt { 2 } } =0.028\times { 10 }^{ -3 }s\)
\(\\ =2.8\times { 10 }^{ -5 }s\)
(d) Force exerted by a rain drop
\(F=\frac { Change \ in \ momentum }{ Time } \)
\(=\frac { \rho -0 }{ t } =\frac { 4.7\times { 10 }^{ -3 } }{ 2.8\times { 10 }^{ -5 } } \approx 168N\)
(e) Radius of the umbrella(R) = \(\frac { 1 }{ 2 } m\)
\(\therefore \ Area \ of \ the \ umbrella(A)=\pi { R }^{ 2 }\)
\(\\ =\frac { 22 }{ 7 } \times { (\frac { 1 }{ 2 } ) }^{ 2 }=\frac { 22 }{ 28 } =\frac { 11 }{ 14 } \approx 0.8{ m }^{ 2 }\)
Number of drops striking the umbrella simultaneously with average of 5cm or 5 x 10^-2m
\(=\frac { 0.8 }{ { (5\times { 10 }^{ -2 }) }^{ 2 } } =320\)
Net force exerted on umbrella 
= 320 x 168 = 53760 N