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

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

Step-by-step 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*3`

`v = 6 m/s`

We need to calculate the height at which the mailbag dropped

Using equation of motion

`H=ut+(1)/(2)at^2`

Put the value into the formula

`H=0+(1)/(2)*2*(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=√(u^2+2gh)`

Put the value into the formula

`v=√(6^2+2*9.8*9)`

`v=14.57\ m/s`

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