Asked by Timothy
A boy throws a stone vertically up to a man standing at a heigt of 2 m above the boy. Of the stone is thrown up with a velocity of 7.0m s what is the velocity of the stone at the instant when it is caught by the man.
A clear explaination is greatly appreciated
A clear explaination is greatly appreciated
Answers
Answered by
drwls
Kinetic energy of M*g*H is lost, with H = 2.0 m.
The initial Velocity is
V1 = 7.0 m/s
The final kinetic energy is
(1/2) M V2^2 = (1/2) M V1^2 - M g H
Cancel out the M's and solve for the final velocity V2.
V2^2 - V1^2 = -2 g H
The initial Velocity is
V1 = 7.0 m/s
The final kinetic energy is
(1/2) M V2^2 = (1/2) M V1^2 - M g H
Cancel out the M's and solve for the final velocity V2.
V2^2 - V1^2 = -2 g H
Answered by
Timothy
Does that give me a negative velocity?
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