A 0.017-kg acorn falls from a position in an oak tree that is 18.5 meters above the ground. Calculate the velocity of the acorn just before it reaches the ground (rounding your answer to the integer) and its kinetic energy when hitting the ground (rounding your answer to the nearest tenth)

The velocity of the acorn just before it reaches the ground is 19 m/s

The kinetic energy when hitting the ground is 3.1 J

Explanation:

Given;

mass of the acorn, m = 0.017 kg

height of fall, h = 18.5 m

Apply the law of conservation of mechanical energy;

mgh = ¹/₂mv²

gh = ¹/₂v²

v² = 2gh

v = √2gh

v = √(2 x 9.8 x 18.5)

v = 19 m/s

Thus, the velocity of the acorn just before it reaches the ground is 19 m/s

Now, determine the kinetic energy when hitting the ground;

K.E = ¹/₂mv²

K.E = ¹/₂(0.017)(19)²

K.E = 3.09 J

K.E = 3.1 J

Therefore, the kinetic energy when hitting the ground is 3.1 J

the force represented by vector a is normal force. if the hockey puck is moving to the right, when it is hit with a significant force by the hockey stick the puck will move off in a different direction. correct on edge.