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A small mailbag is released from a helicopter that is descending steadily at 1.53 m/s. (a) After 4.00 s, what is the speed of the mailbag? v = m/s (b) How far is it below the helicopter? d = m (c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 1.53 m/s?
A small mailbag is released from a helicopter that is descending steadily at 1.53 m/s. (a) After 4.00 s, what is the speed of the mailbag? v = m/s (b) How far is it below the helicopter? d = m (c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 1.53 m/s?
Answer:A small mailbag is released from a helicopter that is descending steadily at 1.53 m/s. Then,a) after 4s, the speed of the mailbox will be, 40.73m/s downwards.b)78.4m downwards.c) v=37.67m/s downward. and d=78.4m downward.Explanation:To find the answer, we need to know more about theequations of uniformly accelerated motion.What are the equations of uniformly accelerated motion?The equations of uniformly accelerated motion under gravity are,How to solve the problem?a) speed of the mailbag after 4s,b) How far it is below the helicopter,c) If the helicopter is rising steadily, then v and d will be,Thus, we can conclude that, the answers to the question are,a) after 4s, the speed of the mailbox will be, 40.73m/s downwards.b)78.4m downwards.c) v=37.67m/s downward. and d=78.4m downward.Learn more about theequation of uniformly accelerated motionhere:brainly.com/question/28044927#SPJ4...