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

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.