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Physics 21 Online
OpenStudy (anonymous):

A body is moving upward with a vel0city of 500ms^-1 what will be the height?

OpenStudy (anonymous):

@AllTehMaffs

OpenStudy (anonymous):

use the relation v = u -gt and v = 0 so calculate t and then use the relation S = ut -.5gt^2

OpenStudy (anonymous):

u = 500m/s v = 0m/s a = -9.81 m/s^2 You can use dibuaman's method, or forget about time and use the 4th equation of motion \[\cancel{v^2}^0 = u^2 + 2ah\] \[h = -\frac{u^2}{2a}\] |dw:1383200487207:dw|

OpenStudy (anonymous):

ty MASTER maff :)

OpenStudy (anonymous):

hahaha, welcome ^_^

OpenStudy (gtxmuqsit):

\[1/2 mv ^{2}=mgh\]

OpenStudy (gtxmuqsit):

by cancelling mass with each other we get\[1/2 v ^{2} = gh \]

OpenStudy (gtxmuqsit):

then we get\[v ^{2}/2g=h\]

OpenStudy (gtxmuqsit):

use the above equation to solve the problem

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