Answer: 121.2 m/s
Explanation:
Here we will use the conservation of energy.
We can write the total energy of an object as:
E = K + U
For this particular case we have:
K = kinetic energy = (m/2)*v^2
U = potential energy = m*g*h
where:
m = mass of the object.
g = gravitational acceleration = 9.8m/s^2
h = height at which the object is dropped, in this case 750m
v = velocity of the object.
Right when the packet is dropped, it's velocity is equal to zero, this means that there is no kinetic energy and only potential energy, then we have:
Ei = Ui = m*9.8m/s^2*750m = m*(7350m^2/s^2)
And when the object is about to hit the ground, h will be almost equal to zero, this will mean that we will have only kinetic energy, and because of the conservation of energy, this final energy must be the same as the initial energy, then we will have
Ei = Ef
m*(7350m^2/s^2) = Ef = Kf
m*(7350m^2/s^2) = (m/2)*v^2
We can divide by m in both sides to get:
(7350m^2/s^2) = (1/2)*v^2
√( 2*(7350m^2/s^2)) = v = 121.2 m/s
Then the velocity of the packet when it reaches the ground is 121.2 m/s
