Acoconut falls out of a tree 12.0 m above the ground and hits a bystander 3.00 m tall on the top of the head. it bounces back up 1.50 m before falling to the ground. if the mass of the coconut is 2.00 kg, calculate the eg of the coconut relative to the ground at each of the following sites: assume no loss due to friction or air resistance.
a. eg while it is still in the tree
b. eg when it hits the bystander on the head
c. eg when it bounces up to its maximum height
d. at what point will it have the most ek?

The energy of the coconut at various heights are required.

235.44 J

58.86 J

88.29 J

is maximum when it hits the bystander.

m = Mass of coconut = 2 kg

g = Acceleration due to gravity =

Distances from the ground

= On the tree = 12 m

= Head of person = 3 m

= Bounced maximum height =

Potential energy in the tree

Bystander head

Bounced maximum height

Kinetic energy is given by

The velocity (v) of the coconut will be maximum after it has traveled the most distance downward.

When the coconut hits the person the distance it traveled was .

After it bounces up and comes down the downward distance traveled is 4.5 m.

Hence, is maximum when it hits the bystander.

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Explanation:

Eg = mgh

a. Eg = (2.00 kg) (9.8 m/s²) (12.0 m)

Eg = 235 J

b. Eg = (2.00 kg) (9.8 m/s²) (3.00 m)

Eg = 58.8 J

c. Eg = (2.00 kg) (9.8 m/s²) (4.50 m)

Eg = 88.2 J

d. When the coconut hits the bystander:

Ek = 235 J − 58.8 J = 176 J

When the coconut hits the ground:

Ek = 88.2 J − 0 J = 88.2 J

Ek is the greatest when the coconut hits the bystander.

true tan speed = ang vel x radius

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