The height of a helicopter above the ground is given by h = 3.0 t^2, where h is in meters and t is in seconds. After 5.0 s, the helicopter releases a small mailbag. How long after its release does the mailbag Preach the ground

3.91 s

Explanation:

From the question,

The height of the helicopter is given as

h = 3t²......................... Equation 1

After 5 seconds, The height of the helicopter is given as

h = 3(5²)

h = 3×25

h = 75 m.

Note: The small mailbag falls under gravity.

Hence,

h = ut + 1/2(gt²)....................... Equation 2

Where u = initial velocity of the mailbag, t = time, h = height, g = acceleration due to gravity of the mail bag.

Given: h = 75 m, u = 0 m/s ( at the maximum height).

Constant: g = 9.8 m/s²

Substitute these values into equation 2

75 = 0(t) +1/2(9.8t²)

solve for t

1/2(9.8t²) = 75

4.9t² = 75

t² = 75/4.9

t² = 15.31

t =

t = 3.91 s

  the unit of the measure =

explanation:

  the property that is quantified by the energy needed per of material to raise the temperature by one.

  si unit of energy = joule (j)

  si unit of temperature = kelvin (k)

  si unit of mass of material = kilogram (kg)

the property that is quantified by the energy needed per of material to raise the temperature by one is called specific heat.

the unit of specific heat is

option number c is the right answer.