A helicopter rises at 4.5 m/s when a bag of its cargo is dropped. After 2.5 s, what is the bag's velocity? ?

How far has the bag fallen? How far below the helicopter is the bag? A helicopter rises at 4.5 m/s when a bag of its cargo is dropped. After 2.5 s, what is the bag's velocity? ?
%26lt;%26lt; After 2.5 s, what is the bag's velocity %26gt;%26gt;





Working formula is





Vf - Vo = gT





where





Vf = final velocity


Vo = initial velocity = 4.5 m/sec.


g = acceleration due to gravity = 9.8 m/sec^2 (constant)


T = 2.5 sec.





Substituting appropriate values,





Vf - (-4.5) = 9.8(2.5)





NOTE the negative sign attached to the initial velocity. This is because the its initial direction (upward) is opposite to the actual direction of the bag (downward).





Solving for Vf,





Vf = 9.8*2.5 - 4.5





Vf = 20 m/sec








%26lt;%26lt; How far has the bag fallen? %26gt;%26gt;





S = VoT + (1/2)(g)T^2





where





S = distance travelled by bag





and all the other items have been previously defined.





Substituting values,





S = -4.5(2.5) + (1/2)(9.8)(2.5)^2





S = 19.375 meters








%26lt;%26lt; How far below the helicopter is the bag? %26gt;%26gt;





ASSUMING that the helicopter is rising at a constant speed, then 2.5 sec after the bag is released, it has travelled 2.5 * 4.5 = 11.25 m.





Therefore, after 2.5 seconds of releasing the bag, the distance between it and the helicopter is





11.25 + 19.375 = 30.625 mA helicopter rises at 4.5 m/s when a bag of its cargo is dropped. After 2.5 s, what is the bag's velocity? ?
here, initial velocity of bag = + 4.5 m/s (up is positive)





to find bag's velocity after 2.5 s, use vf^2 = vi^2 + 2at (substitute a = -9.8 and t = 2.5)





to find how far it's fallen, use d = vi * t + 1/2 a * t ^2 (sub t from above)





to find how far below the helicopter it is, add the distance from above, and the d = v*t (v = 4.5, t = 2.5) of helicopter and there you go.