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

Answer:

In 2.748 sec the mailbag reached the ground

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 $h=3.25t^3=3.25\times 2.25^3=37.01m$

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 $h=ut+\frac{1}{2}gt^2$

$37.01=0\times t+\frac{1}{2}\times 9.8\times t^2$

$t^2=7.553$

t = 2.748 sec