(a) If the package is dropped from a helicopter that is descending steadily at a speed v0, the initial speed of the package is also v0. The acceleration of the package is due to gravity, which is g. We can use the equation:
v = v0 + gt
where v is the speed of the package after t seconds. Therefore, the speed of the package after t seconds is:
v = v0 + gt
(b) To find the distance d, we can use the equation:
d = v0t + (1/2)gt^2
The distance d from the helicopter in terms of g and t is:
d = v0t + (1/2)gt^2
(c) If the helicopter is rising steadily at the same speed, the initial speed of the package would be -v0 (since up is now considered negative).
(a) The speed of the package after t seconds would be:
v = -v0 + gt
(b) The distance d would now be:
d = -v0t + (1/2)gt^2
A package is dropped from a helicopter that is descending steadily at a speed v0. After t seconds have elapsed, consider the following.
(a) What is the speed of the package in terms of v0, g, and t? (Use any variable or symbol stated above as necessary. Let down be positive.)
(b)What distance d is it from the helicopter in terms of g and t?
(c) What are the answers in parts (a) and (b) if the helicopter is rising steadily at the same speed?
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