40. (a) From Table 6-1 and Eq. 6-16, we have vFCAmgvtgt=¡=222ρwherevt= 60 m/s. We estimate the pilot’s mass at about m= 70 kg. Now, we convert v= 1300(1000/3600)≈360 m/s and plug into Eq. 6-14: DCAvmgvvmgvvtt==FHGIKJ=FHGIKJ121222222which yields D= (70 kg)(9.8 m/s2)(360/60)2≈2 ×104N. (b) We assume the mass of the ejection seat is roughly equal to the mass of the pilot.
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