Question

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 1

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 the equations of uniformly accelerated motion.

What are the equations of uniformly accelerated motion?

• The equations of uniformly accelerated motion under gravity are,

                                $v=u+at\\S=ut+\frac{1}{2}at^2\\ v^2-u^2=2aS\\S=\frac{v+u}{2}t$

How to solve the problem?

• a) speed of the mailbag after 4s,

            $v=u+gt=-1.53-(9.8*4)=-40.73 m/s^2$

• b) How far it is below the helicopter,

           $d_1=ut+\frac{1}{2}gt^2=(-1.53*4)+\frac{-9.8*16}{2} =-84.52 m\\d_2=vt=-40.73*4=-162.92 m\\d=-162,92+84.52=-78.4m$

• c) If the helicopter is rising steadily, then v and d will be,

                 $v=u+gt=1.53-(9.8*4)=-37.67 m/s$

                 $d_1=ut+\frac{1}{2}gt^2=(1.53*4)+\frac{-9.8*16}{2} =-72.28 m\\ d_2=vt=-37.67*4=-150.68 m\\d=-150.68+72.28=-78.4 m$

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 the equation of uniformly accelerated motion here:

brainly.com/question/28044927

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