Question

A man descends to the ground from an aeroplane with the help of a parachute which is

hemispherical having a diameter of 4m against the resistance of the air with a uniform velocity

of 25m/s. Find the weight of the man if the weight of the parachute is 10N. Take coefficient of

drag CD= 0.6 and density of air = 1.25kg/m3. 

Answer 1

Explanations & Calculations

• As he descends at a uniform velocity, that means there is no net force on the parachute-man system: all the forces on them are in equilibrium.
• Forces acting on them are the drag force due to the air, the weight of the mas & the weight of the parachute
• Drag force on the man can be neglected compared to that on the parachute
• Drag force is given by

$\qquad\qquad \begin{aligned} \small F_d&=\small \frac{1}{2}C\rho Av^2 \end{aligned}$ . C= drag coefficient

• The average area of the parachute that acts against the air is a circle of radius 2m
• Drag force on the parachute is then,

$\qquad\qquad \begin{aligned} \small F_d &=\small \frac{1}{2}\times(0.6)\times1.25kgm^{-3}\times \pi\cdot2^2\times (25ms^{-1})^2\\ &=\small 2945.24\,N \end{aligned}$

• If the weight of the man is $\small W_m$ then,

$\qquad\qquad \begin{aligned} \small \uparrow F_d&=\small \downarrow ({W_m+W_p})\\ \small 2945.24&=\small W_m+10N\\ \small W_m&=\small \bold{2935.24N} \end{aligned}$