IGNOU MTE 14 2026 - A parachutist, whose weight (actually mass) is , drops from a hel

Question

A parachutist, whose weight (actually mass) is $64\text{ kg}$ , drops from a helicopter $5000\text{ m}$ above the ground. She falls towards the earth under the influence of gravity. Assume that the gravitational force is constant. Assume that the force due to air resistance is proportional to the velocity of the parachutist. The proportionality constant is $k_1 = 16\text{ kg/sec}$ when the parachute is closed, and is $k_2 = 100\text{ kg/sec}$ when it is open. If the parachute does not open until $1\text{ minute}$ after the parachutist leaves the helicopter, after how many seconds will she hit the ground?
b) A projectile is fixed with a constant speed v at two different angles of projection $\alpha$ and $\beta$ such that it gives the same range. Show that $\operatorname{cosec} \alpha = \sec \beta$

09 Jan 2026

Answer :

Word Count : 332

Let downward direction be taken as positive. The mass of the parachutist is (m=64) kg. The gravitational force is (mg=64\times 9.8=627.2) N. The resistive force is proportional to velocity and acts upward, so its magnitude is (kv). While the parachute is closed ((k_1=16) kg/s), the equation of motion is [ m\frac{dv}{dt}=mg-k_1 v. ] This is a first-order linear equation. Solving with the initial condition (v(0)=0), [ v(t)=\frac{mg}{k_1}\left(1-e^{-(k_1/m)t}\right). ] Here (\frac{mg}{k_1}=\frac{627.2}{16}=39.2) m/s and (\frac{k_1}{m}=\frac{16}{64}=0.25\ \text{s}^{-1}). Hence [ v(t)=39.2\left(1-e^{-0.25t}\right). ] The distance fallen in the first phase is [ s_1(t)=\int_0^t v(\tau),d\tau =39.2\left[t+\frac{1}{0.25}\left(e^{-0.25t}-1\right)\right]

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