Question

Instructions help Question 3 (of 10) > Save & Ext Submit 3. value: 10.00 points The Mars roving-laboratory parachute, a 55-ft-diameter disk-gap-band chute with a measured drag coefficient of 1.12. Mars has very low density, about 2.9E-5 slug/ft?, and its gravity is only 38% of earth gravity. If the mass of payload and chute is 1200 kg, estimate the terminal fall velocity of the parachute. Round the final answer to three decimal places. mi/h Hints References eBook & Resources Hint #1 Check my work 2020 McGraw-Hill Education. All rights reserved

Answer 1

Ans) Given,

Mass of parachute (m) = 1200 kg

Diameter of chute (D) = 55 ft or 16.76 m

Density of air ($\rho$) = 2.9 x $10^{-5}$ slugs/ft3 or 0.015 kg/m3

Acceleration due to gravity at mars (g) = 0.38 x 9.81 = 3.73 m/s2

To determine terminal velocity, equate parachute weight by chute drag,

Weight of parachute on mars = m g = 1200 (3.73) = 4476 N  

=> 0.5 Cd $\rho$ A $V^2$ = 4476

where, Cd = drag coefficient = 1.12

A = Area = ($\pi$/4)$D^2$ = ($\pi$/4)($(16.76)^2$ = 220.50 m2

V = Velocity of parachute

Putting values,

=> 0.5 x 1.12 x 0.015 x 220.50 x $V^2$ = 4476

=> $V^2$ = 2416.58

=> V = 49.15 m/s or 110 mi/hr

Hence, terminal velocity of parachute is 110 mi/hr