IGNOU PHE 5 2026 - A parachutist is falling with a speed of when his parachute

Question

A parachutist is falling with a speed of $50 \text{ ms}^{-1}$ when his parachute opens. If the air resistance is (Mv²)/25 where M is the total mass of the man and his parachute, find the speed of the man as a function of time t after the parachute opens. Take $g = 10 \text{ ms}^{-1}$ .

29 Jan 2026

Answer :

Word Count : 466

The problem involves a parachutist falling under gravity with quadratic air resistance. Let’s solve it step by step. Let: * (M) = mass of parachutist with parachute * (v(t)) = velocity at time (t) after parachute opens * (g = 10~\text{m/s²}) * Air resistance (R = \frac{M v^2}{25}) The equation of motion is: [ M \frac{dv}{dt} = Mg - \frac{M v^2}{25} ] Divide through by (M): [ \frac{dv}{dt} = g - \frac{v^2}{25} = 10 - \frac{v^2}{25} ] This is a separable differential equation: [ \frac{dv}{10 - v^2/25} = dt ] Factor (1/25) from the denominator: [ \frac{dv}{10 - \frac{v^2}{25}} = \frac{dv}{\frac{250 - v^2}{25}} = \frac{25 , dv}{250 - v^2} ] So: [

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