A stone is thrown vertically upward and after a sending a height h it comes back to the hand of thrower what is the total distance covered?
Answers
while going up the stone covered a distance of h units
As the stone started from the thrower's hand and ended at the same place.
thus
total distance covered = 2*h
total displacement =0
lets say that the mass of the stone is m
lets say that the mass of the stone is mthe stone is thrown up with an initial velocity u and it reaches a certain height h where its final velocity is 0 m/s.
lets say that the mass of the stone is mthe stone is thrown up with an initial velocity u and it reaches a certain height h where its final velocity is 0 m/s.then the stone falls under the influence of gravity over the same distance.
lets say that the mass of the stone is mthe stone is thrown up with an initial velocity u and it reaches a certain height h where its final velocity is 0 m/s.then the stone falls under the influence of gravity over the same distance.so the kinetic energy of the stone is 1/2mu2
lets say that the mass of the stone is mthe stone is thrown up with an initial velocity u and it reaches a certain height h where its final velocity is 0 m/s.then the stone falls under the influence of gravity over the same distance.so the kinetic energy of the stone is 1/2mu2 the point where v =0 the kinetic energy is converted into potential energy which is mgh
lets say that the mass of the stone is mthe stone is thrown up with an initial velocity u and it reaches a certain height h where its final velocity is 0 m/s.then the stone falls under the influence of gravity over the same distance.so the kinetic energy of the stone is 1/2mu2 the point where v =0 the kinetic energy is converted into potential energy which is mgh so 1/2mu2=mgh
lets say that the mass of the stone is mthe stone is thrown up with an initial velocity u and it reaches a certain height h where its final velocity is 0 m/s.then the stone falls under the influence of gravity over the same distance.so the kinetic energy of the stone is 1/2mu2 the point where v =0 the kinetic energy is converted into potential energy which is mgh so 1/2mu2=mgh which gives h=(u2)/(2g)
lets say that the mass of the stone is mthe stone is thrown up with an initial velocity u and it reaches a certain height h where its final velocity is 0 m/s.then the stone falls under the influence of gravity over the same distance.so the kinetic energy of the stone is 1/2mu2 the point where v =0 the kinetic energy is converted into potential energy which is mgh so 1/2mu2=mgh which gives h=(u2)/(2g) since it comes back to your hand again it has travelled a distance of 2 h so the final expression is h=(u2)/(g)