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Mathematics 21 Online
OpenStudy (aaronandyson):

A ball of mass 8kg is dropped from a height of 10.What is the velocity with which it strikes the ground.

ganeshie8 (ganeshie8):

you may use conservation of energy : \[mgh = \dfrac{1}{2}mv^2 \implies v = \sqrt{2gh}\]

OpenStudy (aaronandyson):

??????????????????????

OpenStudy (aaronandyson):

what is 'g'? @ganeshie8

ganeshie8 (ganeshie8):

\(g\) is acceleration due to gravity

OpenStudy (aaronandyson):

10 ms?

ganeshie8 (ganeshie8):

g = 10 m/s^2 will do http://en.wikipedia.org/wiki/Gravity_of_Earth

OpenStudy (aaronandyson):

solve?

OpenStudy (aaronandyson):

Cant I use mgh = mv^2/2?

OpenStudy (aaronandyson):

@ganeshie8

OpenStudy (aaronandyson):

@ganeshie8

ganeshie8 (ganeshie8):

yes

OpenStudy (aaronandyson):

So now?

ganeshie8 (ganeshie8):

mgh = mv^2/2? cancel m both sides, what do u get ?

OpenStudy (aaronandyson):

10 x 10 = 1/2 x v^2

ganeshie8 (ganeshie8):

yes solve v

OpenStudy (aaronandyson):

50 = v^2?

ganeshie8 (ganeshie8):

nope

OpenStudy (aaronandyson):

200 = v^2

ganeshie8 (ganeshie8):

yes

ganeshie8 (ganeshie8):

take squareroot and finish it off

OpenStudy (aaronandyson):

sqrt200 = v

OpenStudy (aaronandyson):

\[v = \sqrt200\]

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