Question

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

Answer 1

3.91 s

Step-by-step 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 =
`√(15.31)`

t = 3.91 s