A small mailbag is released from a helicopter that isdescending steadily at 1.00

Question

A small mailbag is released from a helicopter that isdescending steadily at 1.00 m/s. (a) After 5.00 s, what is the speed of the mailbag? m/s (b) How far is it below the helicopter? m (c) If the helicopter was rising steadily at 1.00 m/s, whatwould be the results for questions (a) and (b)? A small mailbag is released from a helicopter that isdescending steadily at 1.00 m/s. (a) After 5.00 s, what is the speed of the mailbag? m/s (b) How far is it below the helicopter? m (c) If the helicopter was rising steadily at 1.00 m/s, whatwould be the results for questions (a) and (b)?

Explanation / Answer

a) vi = 1.00m/s vf=vi+gt = 1.00m/s +9.8m/s2*5.00s = 50 m/s b) for the mailbag H1 =vit+(1/2)gt2 = 1.00m/s*5.00s +(1/2)*9.8m/s2*(5.00s)2 = 127.5 m for the helicopter, H2 = vi*5.00s = 5.00m H=127.5 m - 5.00m = 122.5 m so the mailbag is 122.5 m below. c) vi = -1.00m/s vf=vi+gt = -1.00m/s +9.8m/s2*5.00s = 48 m/s for the mailbag H1 =vit+(1/2)gt2 = -1.00m/s*5.00s +(1/2)*9.8m/s2*(5.00s)2 = 117.5 m for the helicopter, H2 = vi*5.00s = -5.00m H=117.5 m - (- 5.00m) = 122.5 m so the mailbag is 122.5 m below. vi = -1.00m/s vf=vi+gt = -1.00m/s +9.8m/s2*5.00s = 48 m/s for the mailbag H1 =vit+(1/2)gt2 = -1.00m/s*5.00s +(1/2)*9.8m/s2*(5.00s)2 = 117.5 m for the helicopter, H2 = vi*5.00s = -5.00m H=117.5 m - (- 5.00m) = 122.5 m so the mailbag is 122.5 m below.

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