Question

44. The height of a helicopter above the ground is givenby, where h is in meters and t is inseconds. After 2.00s, the helicopter releases a small mailbag. Howlong after its release does the mailbag reach the ground? Show allwork!

Answer 1

Height above ground, at t = 2.0 s
                  h = 3.00 t³ = 3.00x8 = 24.0 m

Velocity of the copter, after t = 2.0 s,

v = dh/dt = 9.0t² = 9.0x2x2 = 36.0m/s

height h is increasing withtime. Velocity of the copter andhence that of the mailbag at the time of release are upward;positive.

The net displacement of the mailbag and acceleration ofgravity are downwaard and hence negative. a = g = - 9.8m/s²

          S = ut+ 0.5at²

        S = ut - 4.9t²

      -24 = 36.t - 4.9t²

    4.9.t² - 36.t - 24 =0

     Solving, t = 7.96 s

Answer 2

First of all, you would need to find the height after 2.00seconds:

= 3.00(2.00)³ =3.00(8.00) = 24.0 m

Then you use a formula to find the final velocity before ithit the ground:

V_f² = V_i² +2gd

NOTE: Vf needs to be negative because it is goingdownward.

Vi = 0 m/s because it doesn't say the helicopter was movingup.

g = -9.81 m/s²

d = 24.0 m

-(V_f² )= (0 m/s)² + 2(-9.81m/s²)(24.0 m)

-(V_f²) = -470.88 or-471 m²/s² as significantdigits

Then I can divid by negative to get the number to be positiveso I can find the square root of it.

√(471 m²/s² ) = 21.7 m/s and makeit negative because it's going downward so its -21.7 m/s.

Now I can find the time finally.

Vf =Vi + gt

- 21.7 m/s = 0 m/s + (-9.81 m/s²)(t)

I divid both sides by -9.81 m/s² and I get:

2.21 s = t

Answer 3

Hi!
Ok so starting with:
h = 3.00t³
Also they give us: t = 2.00s
So, plugging in t we get:
h = 3.00(2.00)³
h = 3.00(8.00)
h = 24.00 meters is when the mailbag is released!

What we have is:
distance = 24.00 meters
So we can use the formula: d = v₀* t + (1/2)at²
where v₀ = the initial velocity of the mailbag
   v = the finalvelocity of the mailbag
           a =acceleration acting on the mailbag
           t = time it takes to travel thatdistance

Since the mailbag is being released from the airplane,
there is no PUSH or initial velocity, its just being dropped.

So, we can say
v₀ = 0
So, d = 0 + (1/2) at²
      24.00m = (1/2) (9.81m/s²)(t²)
      48.00m = (9.81m/s²)(t²)
      4.89s² =(t²)                                                   sidenote:s² is just a unit. When you                                                                                                divide 48m with 9.81m/s² the                                                                                           m/s² is the acceleration of                                                                                 gravity (meters per second per second)
    2.21s   =t
Hope this helps!