Solving a Helicopter Height Equation

[rokas]

Homework Statement

The height of a helicopter is given by h=3.00t^3, where h is in meters and t in seconds. After 2s, the helicopter releases a small bag. How long does it take for the bag to reach the ground?

Xi = 0
Xf = 24 (i got 24 meters when i plugged in seconds in h=3.00t^3)
A(gravity) = 9.81
Vi = 0

Homework Equations

Xf=Xi+Vi(t)+1/2at^2

The Attempt at a Solution

i plugged in the numbers

24=(t)+1/2(9.81)(t^2)
24=(t)+4.905(t^2)
19.095=t^3

this is where i am stuck because i don't know how to solve an equation with a letter cubed...
i don't know if I am doing this right, help?

 

[rl.bhat]

Helicopter rising. Its initial velocity in the upward direction is given by dh/dt. Find vi at 2 seconds.
Initial height from the ground xo =24 m. Final height xf = 0. Find t.

 

[tomcue]

I would approach this problem by first clarifying the initial conditions. Is the bag released at the same time as the helicopter takes off, or does it take 2 seconds for the bag to be released after the helicopter takes off? This information is important for accurately solving the problem.

Assuming that the bag is released at the same time as the helicopter takes off, we can use the equation Xf=Xi+Vi(t)+1/2at^2 to solve for the time it takes for the bag to reach the ground. Since the bag is released at the same height as the helicopter, Xi=0. We also know that the bag is initially at rest, so Vi=0. Plugging in these values, we get:

Xf=0+0(t)+1/2(9.81)(t^2)
Xf=4.905t^2

Now, we can plug in the value for Xf from the problem, which is 24 meters:

24=4.905t^2

To solve for t, we can divide both sides by 4.905 and take the square root:

t=√(24/4.905)
t=2.197 seconds

Therefore, it takes approximately 2.197 seconds for the bag to reach the ground after being released from the helicopter. If the bag is released 2 seconds after the helicopter takes off, we would need to adjust the initial conditions and solve the problem again.