A small mailbag is released from a helicopter that is descendingsteadily at 1.50

Question

A small mailbag is released from a helicopter that is descendingsteadily at 1.50 m/s. (a) After 5.00 s, whatis the speed of the mailbag?
Enter anumber. 1 m/s

(b) How far is it below the helicopter?
Enter anumber. 2 m

(c) What are your answers to parts (a) and (b) if the helicopter isrising steadily at 1.50 m/s?
Enter anumber. 3 m/s
Enter anumber. 4 m (a) After 5.00 s, whatis the speed of the mailbag?
Enter anumber. 1 m/s

(b) How far is it below the helicopter?
Enter anumber. 2 m

(c) What are your answers to parts (a) and (b) if the helicopter isrising steadily at 1.50 m/s?
Enter anumber. 3 m/s
Enter anumber. 4 m Enter anumber. Enter anumber. Enter anumber. Enter anumber.

Explanation / Answer

a) let downward be positive v0 = 1.50m/s v(t) = v0 + gt v(5.00s) = 1.50m/s+9.8m/s2 *(5.00s) = 50.5m/s answer: 50.5m/s b) the distance travelled: D = v0 t + (1/2) gt2 = 1.50m/s*5.00s + (1/2) *9.8m/s2*(5.00s)2 = 130.0m but at the same time, helicoper travel down d1 = 1.5m/s*5.00s= 7.500m the distance between the mailbag and helicopter is: H = D-d1 = 122.5m answer: 122.5m c) v0 = -1.50m/s v(t) = v0 + gt v(5.00s) = -1.50m/s+9.8m/s2 *(5.00s) = 47.5m/s answer: 47.5m/s to (a) and: the distance travelled: D = v0 t + (1/2) gt2 = -1.50m/s*5.00s + (1/2) *9.8m/s2*(5.00s)2 = 115.0m but at the same time, helicoper travel down d1 = -1.5m/s*5.00s= -7.500m the distance between the mailbag and helicopter is: H = D-d1 = 122.5m answer: 122.5m to (b) v0 = -1.50m/s v(t) = v0 + gt v(5.00s) = -1.50m/s+9.8m/s2 *(5.00s) = 47.5m/s answer: 47.5m/s to (a) and: the distance travelled: D = v0 t + (1/2) gt2 = -1.50m/s*5.00s + (1/2) *9.8m/s2*(5.00s)2 = 115.0m but at the same time, helicoper travel down d1 = -1.5m/s*5.00s= -7.500m the distance between the mailbag and helicopter is: H = D-d1 = 122.5m answer: 122.5m to (b) the distance travelled: D = v0 t + (1/2) gt2 = -1.50m/s*5.00s + (1/2) *9.8m/s2*(5.00s)2 = 115.0m but at the same time, helicoper travel down d1 = -1.5m/s*5.00s= -7.500m the distance between the mailbag and helicopter is: H = D-d1 = 122.5m answer: 122.5m to (b)

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