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A 0.017-kg acorn falls from a position in an oak tree that is 18.5 meters above the ground.
a. What is the velocity of the acorn just before it reaches the ground?
b. What is the kinetic energy of the acorn when it hits the ground? (Ignore air
resistance.)

Respuesta :

Answer:

Part a)

Final speed of the corn is 19.05 m/s

Part b)

Kinetic energy of the corn is 3.1 J

Explanation:

Part a)

As we know that the initial position of the corn is

h = 18.5 m

now we also know that it will fall from rest and moving under constant acceleration so we will have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]v_f^2 - 0 = 2(9.81)(18.5)[/tex]

[tex]v_f = 19.05 m/s[/tex]

Part b)

Kinetic energy of the corn is given as

[tex]K = \frac{1}{2}mv^2[/tex]

[tex]K = \frac{1}{2}(0.017)(19.05)^2[/tex]

[tex]K = 3.1 J[/tex]

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