Answer to Question #155177 in Mechanics | Relativity for Shalini

Question

A parachutist of mass 70kg falls out of an aeroplane from a height of 2000m and falls under gravity until 600m from the ground when he opens his parachute. The parachute provides a resistance of 2330N. Find the speed at which the parachutist is travelling when he reaches the ground.

Answer 1

Explanations & Calculations

• The speed he attains at a height of 600m is

$\qquad\qquad \begin{aligned} \small V^2&= \small u^2+2as\\ &= \small 0+2\times 9.8\times (2000-600)\\ &=\small 27440m^2s^{-2}\\ \small V&= \small 165.65ms^{-1} \end{aligned}$

• If the parachute provides this resistive force continuously, his acceleration would be

$\qquad\qquad \begin{aligned} \small \downarrow F&= \small ma\\ \small 70kg\times 9.8-2330N&=\small70kg\times a\\ \small a&= \small -23.486ms^{-2} \end{aligned}$

• He experiences a greater deceleration, then if he falls under this deceleration for the rest of the height, then the speed will be

$\qquad\qquad \begin{aligned} \small V_1^2&= \small (165.65ms^{-1})^2+2\times (-23.486ms^{-2})\times 600m\\ \small V_1^2&= \small -746.278 \end{aligned}$

• This means he does not move that way. What happens here is he attains a terminal velocity, that higher deceleration helps him towards that.
• Normally the drag force provided by a parachute varies with the instantaneous speed the parachuter has & it is not a fixed value like 2330N.
• To calculate the terminal speed ou need to know the quantities,

1. The drag coefficient of the parachute & the person
2. The area of the parachute & the person facing the moving direction
3. The density of air

• And the drag force is calculated according to the equation,

$\qquad\qquad \begin{aligned} \small F_d&= \small\frac{1}{2}C\rho A v^2 \end{aligned}$