Question

The height of a helicopter above the ground is given by h = 2.80t3, where h is in meters and t is in seconds. At t = 1.55 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground? ...?

Answer 1

h = 2.80t^3

h = 2.80 (1.55)^3 = 10.43

10.43 = 1/2 gt^2

10.43 x 2/9.8 = t^2

t = √2.12

= 1.45

Hope this helps

Answer 2

The time is 4.692 sec.

Step-by-step explanation:

Given that,

Height
`h = 2.80t^3`

Time t = 1.55 s

We know that,

The rate of change of height is the velocity.

So, the velocity is at t = 1.55 s

`(dh)/(dt)= v = 3*2.80*(1.55)^2`

`v=20.181\ m/s`

The velocity is upward with respect to the ground

We need to calculate the distance above the releasing point

Using equation of motion

`v^2=u^2-2gs`

Put the value into the formula

`s=(v^2)/(2g)`

`s=(20.181^2)/(2*9.8)`

`s=20.77\ m`

The height of the helicopter releases a small mailbag

`h=2.80*(1.55)^3`

`h = 10.43\ m`

We need to calculate the time

Using equation of motion

`s=ut-(1)/(2)gt^2+h`

Put the value into the formula

`0=20.77* t-(1)/(2)*9.8* t^2+10.43`

`t=-0.454,4.692`

On neglecting negative value of time

Hence, The time is 4.692 sec.