If an object is thrown directly upwards (initial speed 3m/s) what will its maximum height be?

Question

anonymous · Answer

oh, and the only other force acting on it is gravity (9.8m/s)

ganeshie8 · Answer

there are several ways to do this - my fav is conservation of energy :

inital PE + KE   =  final PE + KE

anonymous · Answer

what is PE and KE

ganeshie8 · Answer

good question,
PE = Potential Energy 
KE = Kinetic Energy

ganeshie8 · Answer

since initially there was no PE, 
and at maximum height there wont be any KE, the equation above simplifies to :

initial KE = final PE

ganeshie8 · Answer

$\large \dfrac{1}{2}m v_i^2   =  mgh$

ganeshie8 · Answer

$\large \dfrac{1}{2} v_i^2   =  gh$


$\large \dfrac{v_i^2}{2g}    =  h$

ganeshie8 · Answer

plugin the values and evaluate ^

anonymous · Answer

so $$v _{i} $$ is 2m/s

ganeshie8 · Answer

you're given $\large   v_i = 3m/s$ right ?

anonymous · Answer

oh, sorry...yep I meant that

anonymous · Answer

haha!! If I plug those values in I get h=0.2093 which is very small?

anonymous · Answer

wait... h=0.469

anonymous · Answer

that's still really small...

ganeshie8 · Answer

throw the stone bit faster if u want it go high :)

anonymous · Answer

oh...so that's right?

anonymous · Answer

can I also ask in the first equation you gave what does m mean?

ganeshie8 · Answer

yep u get that much only for that speed lol :P

ganeshie8 · Answer

anonymous · Answer

$$mass \times gravity \times height$$

anonymous · Answer

ok thanks :)

anonymous · Answer

No problem