Question

Hi there, it is your average pathetic Junior. Anyways, I need help on 8&9 ASAP, this assignment is beyond overdue but hey who cares. Oh yeah me, which is why I am asking for help. SO please give it to me :). Show all work please. Giving brainliest.

Answer 1

Answer:

a) -31.36 m/s

b) 50.176 m

Explanation:

a) Velocity of the bag

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)

Hence:

$V_{f}=-31.36m/s$  Note the negative sign is because the direction of the bag is downwards as well.

b) Final height of the bag

In this case we will use the following equation:

$y=V_{o}t-\frac{1}{2}gt^{2}$ (3)

Where:

$y$ is the distance the bag has fallen

$V_{o}=0$ remembering the bag was dropped

$g=-9.8m/s^{2}$ is the acceleration due gravity (downwards)

$t=3.2 s$ is the time

Then:

$y=-\frac{1}{2}gt^{2}$ (3)

$y=-\frac{1}{2}(-9.8m/s^{2})(3.2)^{2}$ (4)

Finally:

$y=50.176 m$