The descent vehicle weighing 100 kg approaches the Earth at a speed of 72 km / h.

The descent vehicle weighing 100 kg approaches the Earth at a speed of 72 km / h. What is the largest impulse of force that the soft-landing engines must impart to the vehicle in order to reduce its landing speed to 1 m / s?

To find the required impulse of force, which must be communicated to the descent vehicle, we will use the formula: I = F * Δt = ΔP = m * V1 – m * V2 = m * (V1 – V2).

Variables: m is the mass of the apparatus (m = 100 kg); V1 is the initial speed of the vehicle (V1 = 72 km / h = 20 m / s); V2 is the required landing speed (V2 = 1 m / s).

Calculation: I = m * (V1 – V2) = 100 * (20 – 1) = 1900 kg * m / s.

Answer: The descent vehicle must be imparted with a force impulse of 1900 kg * m / s.