Asked by anonymous
A volleyball is bumped vertically up to a height of 10 m above the player’s arms. What was the initial velocity of the ball just after leaving the player’s arms?
Answers
Answered by
Damon
(1/2) m v^2 = m g h
v^2 = 2 g h
v^2 = 2 g h
Answered by
Anonymous
-196
Answered by
Hakuho
Kinetic Energy = Gravitational Potential Energy
(0.5)*mass*velocity^2 = mass*acceleration due to gravity*height above gorund
simplify by removing mass on either side => 0.5*v^2=g*h
divide by 0.5 and then square both sides to be left with => v=sqrt[2*g*h]
(0.5)*mass*velocity^2 = mass*acceleration due to gravity*height above gorund
simplify by removing mass on either side => 0.5*v^2=g*h
divide by 0.5 and then square both sides to be left with => v=sqrt[2*g*h]
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