A 250.0 kg rock falls off a 40.0 m cliff. What is the kinetic energy of the rock just before it hits the ground (hint: conservation of energy)?

kinetic energy body when it hits the ground is 98000 joule

1000joule = 1 kilojoule

, kinetic energy body when it hits the ground is 98  kilo-joule

Step-by-step explanation:

conservation of energy states that total energy of a system remains constant.

Potential energy of body = mgh

m = mass

g = gravitational pull = 9/8 m/s^2

h = height

kinetic energy = 1/2 mv^2

where v is the velocity of body

Total energy for this at any point is sum of potential energy and kinetic energy

total energy at height h

v= 0

PE = 250*9.8*40= 98,000

KE = 1/2 m0^2  = 0

total energy at when ball hits the ground

h=0

PE = 250*9.8*0 =

KE = 1/2 mv^2  

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Applying conservation of energy

Total energy at height h = total energy at ground

98000 = KE

Thus, kinetic energy body when it hits the ground is 98000 joule

1000joule = 1 kilojoule

, kinetic energy body when it hits the ground is 98  kilo-joule

based on context clues you gave i'm assuming the answer would be c.

answer is d-29

step-by-step explanation:

since you didn't include the picture, i can give you the description so you know how to pick the correct answer. the third quartile of a data is always shown in a box plot. in the middle of your graph should be a box. the third quartile is the value that goes with the far right edge of the box.