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what do we assume the displacement of the starting position is?

i.e. I came across this question a stone is released from rest 22.5 m above the ground and falls vertically before hitting the ground. calculate the velocity of the stone as it hits the ground.

so what do you set the displacement equal to as it hits the ground. -22.5 or 0? because on a previous thread someone said use it in relation to the ground always but if you use 0 you just get 0 but if you use -22.5 you get an answer of 21?
Original post by jonjoshelvey21
i.e. I came across this question a stone is released from rest 22.5 m above the ground and falls vertically before hitting the ground. calculate the velocity of the stone as it hits the ground.

so what do you set the displacement equal to as it hits the ground. -22.5 or 0? because on a previous thread someone said use it in relation to the ground always but if you use 0 you just get 0 but if you use -22.5 you get an answer of 21?


Depends what you take as your reference point. If you take it to be the ground, then you have displacement of your stone as s=22.5+ut+at2s=22.5+ut+at^2 hence you are looking for the point where s=0s=0 (meaning it has a displacement of 0 from your reference point)

If you take the reference point to be the point you throw it from, then s=ut+at2s=ut+at^2 and you are looking for when the stone is displaced 22.5m metres below the starting point, hence you set s=22.5s=-22.5

In either case, you end up with the same equation to solve.
(edited 6 years ago)
Original post by RDKGames
Depends what you take as your reference point. If you take it to be the ground, then you have displacement of your stone as s=22.5+ut+at2s=22.5+ut+at^2 hence you are looking for the point where s=0s=0 (meaning it has a displacement of 0 from your reference point)

If you take the reference point to be the point you throw it from, then s=ut+at2s=ut+at^2 and you are looking for when the stone is displaced 22.5m metres below the starting point, hence you set s=22.5s=-22.5

In either case, you end up with the same equation to solve.

thank you for the reply but how do you end upneih the same equation to solve if equal displacement to 0 in one equation and to -22.5 in another ?
Original post by jonjoshelvey21
thank you for the reply but how do you end upneih the same equation to solve if equal displacement to 0 in one equation and to -22.5 in another ?


First case, we said we set s=0s=0, so this yields 0=22.5+ut+at20=22.5+ut+at^2 which is the same as 22.5=ut+at2-22.5=ut+at^2

Second case, we set s=22.5s=-22.5 which gives us 22.5=ut+at2-22.5=ut+at^2

Clearly, we get the same equation.

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