Posted by **rose** on Tuesday, October 5, 2010 at 1:51pm.

hello, I need some help with this work problem that goes... If the height of a helicopter above the ground is given by h=3.25t^3, where h is in meters and t is in sec. After 1.50s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

So, to approach this question, I first 1.50s in the formula for h given to get my height in which I got 10.97m... then i am using the formula x=v(i)t-1/2gt^2 where my x is -10.97, the v is 0 and g=8.91 when a substitute all my values to the equation i and up with 2.24=t^2 in which then i square root both sides to give and answer of about 1.49s... I thought that my calculation were right but it seems that the answer is incorrect... please let me know what i am doing wrong... thank you

- physics -
**bobpursley**, Tuesday, October 5, 2010 at 2:48pm
You need the velocity of the helicopter when the bag is dropped.

h=3.25t^3

dh/dt= 9.75 t^2= vi

h(1.5)=3.25*1.5^3=10.97

Now, for the bag:

hf=hi+vit-4.9t^2

0=10.97+9.75(1.5^2)t-4.9t^2

Now, solve that quadratic.

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