Question

A supply bag is dropped from a rescue plane. After the bag falls for 3.2 seconds , what is the velocity of the bag?

Answer 1

Answer: -31.36 m/s

Step-by-step explanation:

This is a problem of motion in one direction (specifically vertical motion), and the equation that best fulfills this approach is:

`V_(f)=V_(o)+a.t` (1)

Where:

`V_(f)` is the final velocity of the supply bag

`V_(o)=0` is the initial velocity of the supply bag (we know it is zero because we are told it was "dropped", this means it goes to ground in free fall)

`a=g=-9.8m/s^(2)` is the acceleration due gravity (the negtive sign indicates the gravity is downwards, in the direction of the center of the Earth)

`t=3.2s` is the time

Knowing this, let's solve (1):

`V_(f)=0+(-9.8m/s^(2))(3.2s)` (2)

Finally:

`V_(f)=-31.36m/s` Note the negative sign is because the direction of the bag is downwards as well.