Question

A man, 65 kg, descends to the ground with the helps of a parachute, 18 kg. The parachute is hemispherical in shape, 2 m diameter. Density of air can be taken as $0.00125g/c{m}^3$ and its kinematic viscosity as 0.15 stoke. What is the terminal velocity of the parachute? [Take $C_D,=1.5andg=1000cm/se{c}^2]$

• A. 16.6 m/sec
• B. 15.8 m/sec
• C. 15.0m/sec
• D. 14.1 m/sec

Answer 1

The correct option is A 16.6 m/sec

$Mg=F_D$

$10\times (65+18)=C_D\times \frac{1}{2}\times \rho \times A\times V_T^2$

$10\times 83=1.5\times \frac{1}{2}\times 1.25\times \frac{\pi }{4}(2{)}^2\times V_T^2$

$V_T=16.79,m/sec$