Question

A helicopter accelerates vertically from the ground from rest at 2 m/s2. 3 s after the helicopter leaves the ground a mailbag is dropped from the helicopter. What is the speed of the mailbag just before it hits the ground? (The helicopter is moving when the bag is dropped.)

Answer 1

Answer:

The velocity of the mailbag just before it hits the ground is 14.57 m/s.

Explanation:

Given that,

Acceleration = 2 m/s

Time = 3 sec

We need to calculate the velocity of mailbag

Using equation of motion

$v=u+at$

Put the value into the formula

$v=0+2\times3$

$v = 6 m/s$

We need to calculate the height at which the mailbag dropped

Using equation of motion

$H=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$H=0+\dfrac{1}{2}\times2\times(3)^2$

$H=9\ m$

We need to calculate the velocity of the mailbag just before it hits the ground

Using equation of motion

$v^2= u^2+2gh$

$v=\sqrt{u^2+2gh}$

Put the value into the formula

$v=\sqrt{6^2+2\times9.8\times9}$

$v=14.57\ m/s$

Hence, The velocity of the mailbag just before it hits the ground is 14.57 m/s.