# Physics Question (1 Viewer)

#### axwe7

##### Member
An astronaut on the moon drops a golf ball to determine the acceleration due to gravity. The time taken to fall through 1.2 m is 1.2 s. Determine the acceleration due to gravity on the moon.

I know that a = (v-u)/t

However the numerator states values based on velocity. Hence we should first use - V = s/t

s = 1.2
t = 1.2

Therefore Velocity = 1 ms-1, initial velocity @ astronaut = 0ms-1.

Therefore,
a = (1-0)/1.2
= 0.8333...

It's incorrect...

What am I doing wrong and what is method should be used to answer this question?

Regards,
Axwe7.

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#### Nailgun

##### Cole World
$\bg_white s=ut+\frac { 1 }{ 2 } a{ t }^{ 2 }\\ \frac { 2(s-ut) }{ { t }^{ 2 } } =a\\ u=0,\quad s=1.2,\quad t=1.2\\ a=\frac { 2.4 }{ { 1. }2^{ 2 } } =1.66667\quad m{ s }^{ -2 } down$

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#### axwe7

##### Member
$\bg_white s=ut+\frac { 1 }{ 2 } a{ t }^{ 2 }\\ \frac { 2s-ut }{ { t }^{ 2 } } =a\\ u=0,\quad s=1.2,\quad t=1.2\\ a=\frac { 2.4 }{ { 1. }2^{ 2 } } =1.66667\quad m{ s }^{ -2 }$
Why can't I use the formula that I used before?

Thanks btw dude.

P.S Isn't it supposed to be s^2-2ut?

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#### Nailgun

##### Cole World
$\bg_white v=\frac { s }{ t } is more accurately { v }_{ avg }=\frac { s }{ t } and so is not actually the final velocity of the golf ball, but the average velocity of the flight$

This link derives Newton's equations http://www.hitxp.com/phy/cph/071202.htm and if you have any trouble understanding I can go through it

#### Nailgun

##### Cole World
$\bg_white These are typically the equations used in preliminary motion. Basically work out what variables you have and sub in. These are revisited and modified in HSC Projectile motion so it would be a good idea to get familiar with these. v=u+at\\ { v }^{ 2 }={ u }^{ 2 }+2as\\ s=ut+\frac { 1 }{ 2 } a{ t }^{ 2 }$

EDIT:
P.S Isn't it supposed to be s^2-2ut?
um no lol

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#### InteGrand

##### Well-Known Member
$\bg_white s=ut+\frac { 1 }{ 2 } a{ t }^{ 2 }\\ \frac { 2s-ut }{ { t }^{ 2 } } =a\\ u=0,\quad s=1.2,\quad t=1.2\\ a=\frac { 2.4 }{ { 1. }2^{ 2 } } =1.66667\quad m{ s }^{ -2 } down$
$\bg_white \noindent The rearranged form is a = \frac{2(s-ut)}{t^2}. Since you got the answer right (according to the given answers), I suppose you just forgot to write the brackets?$

#### Nailgun

##### Cole World
$\bg_white \noindent The rearranged form is a = \frac{2(s-ut)}{t^2}.$
Oh whoopsie
Well, you'd get the same answer because u=0, but yeah lol

#### InteGrand

##### Well-Known Member
Oh whoopsie
Well, you'd get the same answer because u=0, but yeah lol
Ah right (didn't bother to read the actual question haha).

#### Nailgun

##### Cole World
Where did your explanations go ahaha?

#### axwe7

##### Member
Oh whoopsie
Well, you'd get the same answer because u=0, but yeah lol
So it's supposed to be 2ut?

#### InteGrand

##### Well-Known Member
So it's supposed to be 2ut?
Yeah, just an algebraic rearrangement. It made no difference here though as u was just 0, as Nailgun noted.

#### Nailgun

##### Cole World
$\bg_white When the numerator is expanded you would get a=\frac { 2s-2ut }{ { t }^{ 2 } } which is similar to what you said (sorry ahahha), but not { s }^{ 2 }-2ut$

#### axwe7

##### Member
$\bg_white When the numerator is expanded you would get a=\frac { 2s-2ut }{ { t }^{ 2 } } which is similar to what you said (sorry ahahha), but not { s }^{ 2 }-2ut$
NP dude, you still nailed it you gun

See whet I d1d Thair?

#### axwe7

##### Member
Yeah, just an algebraic rearrangement. It made no difference here though as u was just 0, as Nailgun noted.
Thanks by the way

You always seem to help, not matter what.

#### axwe7

##### Member
$\bg_white These are typically the equations used in preliminary motion. Basically work out what variables you have and sub in. These are revisited and modified in HSC Projectile motion so it would be a good idea to get familiar with these. v=u+at\\ { v }^{ 2 }={ u }^{ 2 }+2as\\ s=ut+\frac { 1 }{ 2 } a{ t }^{ 2 }$

EDIT:

um no lol
Thanks man