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:
[tex]V_{f}=V_{o} +a.t[/tex] (1)
Where:
[tex]V_{f}[/tex] is the final velocity of the supply bag
[tex]V_{o}=0[/tex] 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)
[tex]a=g=-9.8m/s^{2}[/tex] is the acceleration due gravity (the negtive sign indicates the gravity is downwards, in the direction of the center of the Earth)
[tex]t=3.2s[/tex] is the time
Knowing this, let's solve (1):
[tex]V_{f}=0+(-9.8m/s^{2})(3.2s)[/tex] (2)
Hence:
[tex]V_{f}=-31.36m/s[/tex] 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:
[tex]y=V_{o}t-\frac{1}{2}gt^{2}[/tex] (3)
Where:
[tex]y[/tex] is the distance the bag has fallen
[tex]V_{o}=0[/tex] remembering the bag was dropped
[tex]g=-9.8m/s^{2}[/tex] is the acceleration due gravity (downwards)
[tex]t=3.2 s[/tex] is the time
Then:
[tex]y=-\frac{1}{2}gt^{2}[/tex] (3)
[tex]y=-\frac{1}{2}(-9.8m/s^{2})(3.2)^{2}[/tex] (4)
Finally:
[tex]y=50.176 m[/tex]
