Kinematics - finding time given height and acceleration

[indietro]

Homework Statement

The height of a helicopter above the ground is given by h = 2.50t^3 , where h is in meters and t is in seconds. After 2.45 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

Homework Equations

d=v1t +1/2at^2

The Attempt at a Solution

So i found the height of the helicopter at 2.45s which also = the distance the bag must fall:
=2.5(2.15)^3
= 36.8 m

then since the bag is in free fall it has an acceleration of 9.8m/s^2
and i just assume that the velocity is 0 because it is being drop from the helicopter.
so using kinematic equation: d=v1t +1/2at^2
rearrange for time and plugging in:
t^2= 36.8/4.9
t = 2.74s

can someone just check that my reasoning is correct? that would be great thanks :)

 

[Redbelly98]

Well, since the helicopter is moving, the bag will have a (non-zero) initial velocity. So if you can figure out the helicopter's velocity at 2.45s, then you can use that for v1.

 

[indietro]

im not sure if I am finding velocity correctly...
so d = 36.76 m
a = 9.8 m/s^2
t = 2.45 s
plug into kinematic equation: d= vt + 1/2at^2 = 3 m/s

or do i not use gravity as acceleration because the helicopter could be accelerating...??

 

[Redbelly98]

The helicopter is rising, not falling with gravity. Also, its acceleration is not constant; since a is not constant, those equations for constant acceleration do not work for the helicopter.

Have you had calculus? The derivative would be helpful in figuring out the helicopter's velocity here.

 

[indietro]

ok so the velocity of the helicopter h'= 7.5t^2 , plugging in 2.45s means the bag has in initial velocity of 45.0 m/s ...?

 

[Redbelly98]

Yes.