Question

Initial small parachute Large parachute deployed at time Mass m Figure 9.34: Prototype dual parachute system.

Answer 1

Given that, sky diver mass = m linear drug force = F_g vertical velocity = v Given, F_B proportionalto v Given first a small parachute open therefore ma = mg - F_B = mg - cf times 1/2 PAV^2 Here, A = projected area V = velocity And Now the second parachute open therefore m dV/dt = m_g - 10 c_f times 1/2 PAV^2 max. velocity when M_g - c_f times 1/2 PAV^2_max = 0 therefore V_max = (2mg/cf PA)^v_2 therefore m dV/dt = mg - cf PAV^2/2 therefore Integral^V_max_0 m/mg - cf PV^2/2 dV = integral _0 dt therefore squareroot 2 mg/2g squareroot c_p PA ln 1 + k times 0.98 V_max/1 - k times 0.98 V_max = t \r\n velocity = 0.98 times squareroot 2mg/c_p PA terminal velocity = V_f therefore m_g - 10c_f/2 PAV^2_f = 0 therefore V_R = (mg/5 c_p PA) = 0. therefoe Integral^V_t_0.98 V_max m dV/mg - 5 c_f PAV^2 = Integral^t_1_t dt therefore 1/2 squareroot 10 kg