Question

The height of a helicopter above the ground is given by h = 3.25t3, where h is in meters and t is in seconds. At t = 2.25 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

_______ s

Answer 1

In 2.748 sec the mailbag reached the ground

Step-by-step explanation:

We have given height from the ground
`h=3.25t^3`

At t =2.25 sec helicopter releases a small mailbag so at t = 2.25 sec height from the ground
`h=3.25t^3=3.25* 2.25^3=37.01m`

When the mail box is drooped its initial velocity would zero so u = 0 m/sec

Acceleration due to gravity
`g=9.8m/sec^2`

According to third law of motion
`h=ut+(1)/(2)gt^2`

`37.01=0* t+(1)/(2)* 9.8* t^2`

`t^2=7.553`

t = 2.748 sec