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

Answer:

The time is 4.692 sec.

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

$\dfrac{dh}{dt}= v = 3\times2.80\times(1.55)^2$

$v=20.181\ m/s$

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

$s=\dfrac{v^2}{2g}$

$s=\dfrac{20.181^2}{2\times9.8}$

$s=20.77\ m$

The height of the  helicopter releases a small mailbag

$h=2.80\times(1.55)^3$

$h = 10.43\ m$

We need to calculate the time

Using equation of motion

$s=ut-\dfrac{1}{2}gt^2+h$

Put the value into the formula

$0=20.77\times t-\dfrac{1}{2}\times9.8\times t^2+10.43$

$t=-0.454,4.692$

On neglecting negative value of time

Hence, The time is 4.692 sec.