physics free fall question?

Question

The height of a helicopter above the ground is given by h = 3.00t^3, where h is in meters and t is in seconds. After 2.00 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

the answer in the book is 7.96 seconds, but can someone explain how to get that?

Answer 1

First solve for the height of release
h=24 meters

then the velocity at the moment of release
v(t)=9*t^2
v(2)=36 m/s upward

Once the bag is released
y(t)=24+36*t-.5*9.81*t^2

solve for y(t)=0
t=7.96

j

Answer 2

when the bag is released, the acceleration is g=9.8 m/s^2 (gravity is the only force acting on the mailbag)
its height is h(2)=3(2)^3=24 m
its velocity is h'(2)=9t^2 |(@t=2) = 36 m/s
(i.e. velocity is dh/dt)

you have the initial conditions, plug them into the equation and solve for t

s-s0=(1/2)*a*t^2 + v0*t
where you must be careful with sign convention
if we take a as +9.8m/s, then v0 must be -36 m/s
that is, downward is the positive direction
and let the point where the mailbag is released be the origin for our calculation, then s0 is 0 and s is +24

solve(4.9*t^2-36*t=24,t);
gives
-.6151593589, 7.962098134
ignore the negative answer
hence t=7.96 sec