Physics Question (1 Viewer)

axwe7 · Feb 5, 2016

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...

The answer is 1.6ms-1.

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

Regards,
Axwe7.

Nailgun · Feb 5, 2016

axwe7 · Feb 5, 2016

Nailgun said:
$s=ut+\frac { 1 }{ 2 } a{ t }^{ 2 }\\ \frac { 2s-ut }{ { t }^{ 2 } } =a\\ u=0,\quad s=1.2,\quad t=1.2\\ a=\frac { 2.4 }{ { 1. }2^{ 2 } } =1.66667\quad m{ s }^{ -2 }$

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?

Nailgun · Feb 5, 2016

$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 · Feb 5, 2016

$$ 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:

axwe7 said:
P.S Isn't it supposed to be s^2-2ut?

um no lol

InteGrand · Feb 5, 2016

Nailgun said:
$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 $$

$\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 · Feb 5, 2016

InteGrand said:
$\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 · Feb 5, 2016

Nailgun said:
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 · Feb 5, 2016

Where did your explanations go ahaha?

axwe7 · Feb 5, 2016

Nailgun said:
Oh whoopsie
Well, you'd get the same answer because u=0, but yeah lol

So it's supposed to be 2ut?

InteGrand · Feb 5, 2016

axwe7 said:
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 · Feb 5, 2016

axwe7 · Feb 5, 2016

Nailgun said:
$$ 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 · Feb 5, 2016

InteGrand said:
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 · Feb 5, 2016

Nailgun said:
$$ 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

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