a stone is thrown vertically upwards. when stone is at a height half of its maximum height it's speed is 10 m/s . what is the maximum height attained by the stone.
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this is the solution bro
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mahendrachoudhary123:
why did you take inittial velocity 20 m/s
Hmm, i did that mistake. wait a second, I am solving it and will send photo of solution within some time
no need. i have already solved.
i have sent it
Answered by
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energy conservation law
at height h/2 total energy=kinetic energy+potential energy
E=1/2m*10*10+m*g*h/2
since kin.enrg.=1/2*mv^2
potent.enrg.=m*g*height
at maximum height kinetic energy is 0 since velocity is 0. so all the energy is potential energy .
E=m*g*h
equating both equations
1/2*m(10*10+g*h)=m*g*h
so m cancels out.
100+g*h=2*g*h
100=g*h
then h=100/g
if g=10m/s^2
h=10m
i could have solved the question by equations of motion but it is easier than that.
you look to be 11 student so may be you are not taught enrgy conservation. so if there is any case then tell me i'll solve by that method too.
at height h/2 total energy=kinetic energy+potential energy
E=1/2m*10*10+m*g*h/2
since kin.enrg.=1/2*mv^2
potent.enrg.=m*g*height
at maximum height kinetic energy is 0 since velocity is 0. so all the energy is potential energy .
E=m*g*h
equating both equations
1/2*m(10*10+g*h)=m*g*h
so m cancels out.
100+g*h=2*g*h
100=g*h
then h=100/g
if g=10m/s^2
h=10m
i could have solved the question by equations of motion but it is easier than that.
you look to be 11 student so may be you are not taught enrgy conservation. so if there is any case then tell me i'll solve by that method too.
above solution looks to be wrong.
Ya mate
was my solution wrong??
no. that's right.
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