How Long Does It Take for a Dropped Packet to Hit the Ground?

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A packet dropped from a helicopter moving upward at 8 m/s from a height of 122.8 m needs to be analyzed using kinematic equations. The relevant equations include vf = vi + gt, S = vit + 1/2gt^2, and 2aS = vf^2 - vi^2. The initial velocity (vi) of the packet is 8 m/s upward, and the distance (S) to the ground is 122.8 m. The discussion focuses on determining which equation to use to solve for the time (t) it takes for the packet to hit the ground. The solution requires showing the work and calculations based on the chosen equation.
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Homework Statement



A Helicopter is vertically moving upward with speed of 8m/s.At a height of 122.8m from ground,a packet is fallen from the Heli window.How much time would the packet take for hitting to ground?

Homework Equations



vf=vi+gt.
S=vit+1/2gt^2
2aS=vf^2-vi^2

The Attempt at a Solution

 
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You have the initial speed of the packet, also the distance from the ground, and you are required to find t,
Which of the three equations you going to use?
a.) vf=vi+gt.
b.) S=vit+1/2gt^2
c.) 2aS=vf^2-vi^2
 
shehri said:

Homework Statement



A Helicopter is vertically moving upward with speed of 8m/s.At a height of 122.8m from ground,a packet is fallen from the Heli window.How much time would the packet take for hitting to ground?

Homework Equations



vf=vi+gt.
S=vit+1/2gt^2
2aS=vf^2-vi^2

The Attempt at a Solution


Well... what have you tried? Show your work please.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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